feat(chapter_backtracking): Add js and ts codes for chapter 13.3 (#667)
* feat(chapter_dynamic_programming): Add js and ts codes for chapter 14.1 * style(chapter_dynamic_programming): format code * refactor(chapter_dynamic_programming): remove main definition and add type * feat(chapter_backtracking): Add js and ts codes for chapter 13.3 * feat(chapter_divide_and_conquer): Add js and ts codes for chapter 12.2 * feat(chapter_divide_and_conquer): Add js and ts codes for chapter 12.3 * feat(chapter_divide_and_conquer): Add js and ts codes for chapter 12.4 * style(chapter_divide_and_conquer): fix typo * refactor: Use === instead of == in js and tspull/664/head^2
William Yuan
1 year ago
committed by
GitHub
parent
c7c33f19ac
commit
70784a1ec3
@ -0,0 +1,46 @@
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/**
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* File: subset_sum_i.js
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 回溯算法：子集和 I */
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function backtrack(state, target, choices, start, res) {
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// 子集和等于 target 时，记录解
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if (target === 0) {
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res.push([...state]);
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return;
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}
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// 遍历所有选择
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// 剪枝二：从 start 开始遍历，避免生成重复子集
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for (let i = start; i < choices.length; i++) {
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// 剪枝一：若子集和超过 target ，则直接结束循环
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// 这是因为数组已排序，后边元素更大，子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 尝试：做出选择，更新 target, start
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state.push(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res);
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// 回退：撤销选择，恢复到之前的状态
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state.pop();
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}
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}
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/* 求解子集和 I */
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function subsetSumI(nums, target) {
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const state = []; // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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const start = 0; // 遍历起始点
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const res = []; // 结果列表（子集列表）
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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const nums = [3, 4, 5];
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const target = 9;
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const res = subsetSumI(nums, target);
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console.log(`输入数组 nums = ${JSON.stringify(nums)}, target = ${target}`);
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console.log(`所有和等于 ${target} 的子集 res = ${JSON.stringify(res)}`);
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@ -0,0 +1,51 @@
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/**
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* File: subset_sum_ii.js
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 回溯算法：子集和 II */
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function backtrack(state, target, choices, start, res) {
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// 子集和等于 target 时，记录解
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if (target === 0) {
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res.push([...state]);
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return;
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}
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// 遍历所有选择
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// 剪枝二：从 start 开始遍历，避免生成重复子集
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// 剪枝三：从 start 开始遍历，避免重复选择同一元素
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for (let i = start; i < choices.length; i++) {
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// 剪枝一：若子集和超过 target ，则直接结束循环
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// 这是因为数组已排序，后边元素更大，子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
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if (i > start && choices[i] === choices[i - 1]) {
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continue;
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}
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// 尝试：做出选择，更新 target, start
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state.push(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i + 1, res);
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// 回退：撤销选择，恢复到之前的状态
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state.pop();
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}
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}
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/* 求解子集和 II */
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function subsetSumII(nums, target) {
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const state = []; // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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const start = 0; // 遍历起始点
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const res = []; // 结果列表（子集列表）
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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const nums = [4, 4, 5];
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const target = 9;
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const res = subsetSumII(nums, target);
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console.log(`输入数组 nums = ${JSON.stringify(nums)}, target = ${target}`);
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console.log(`所有和等于 ${target} 的子集 res = ${JSON.stringify(res)}`);
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/**
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* File: binary_search_recur.js
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 二分查找：问题 f(i, j) */
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function dfs(nums, target, i, j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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const m = i + ((j - i) >> 1);
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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function binarySearch(nums, target) {
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const n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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/* Driver Code */
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const target = 6;
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const nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35];
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// 二分查找（双闭区间）
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const index = binarySearch(nums, target);
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console.log(`目标元素 6 的索引 = ${index}`);
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/**
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* File: build_tree.js
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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const { printTree } = require('../modules/PrintUtil');
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const { TreeNode } = require('../modules/TreeNode');
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/* 构建二叉树：分治 */
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function dfs(preorder, inorder, hmap, i, l, r) {
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// 子树区间为空时终止
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if (r - l < 0) return null;
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// 初始化根节点
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const root = new TreeNode(preorder[i]);
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// 查询 m ，从而划分左右子树
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const m = hmap.get(preorder[i]);
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// 子问题：构建左子树
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root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1);
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// 子问题：构建右子树
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root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r);
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// 返回根节点
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return root;
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}
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/* 构建二叉树 */
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function buildTree(preorder, inorder) {
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// 初始化哈希表，存储 inorder 元素到索引的映射
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let hmap = new Map();
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for (let i = 0; i < inorder.length; i++) {
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hmap.set(inorder[i], i);
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}
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const root = dfs(preorder, inorder, hmap, 0, 0, inorder.length - 1);
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return root;
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}
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/* Driver Code */
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const preorder = [3, 9, 2, 1, 7];
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const inorder = [9, 3, 1, 2, 7];
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console.log('前序遍历 = ' + JSON.stringify(preorder));
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console.log('中序遍历 = ' + JSON.stringify(inorder));
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const root = buildTree(preorder, inorder);
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console.log('构建的二叉树为：');
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printTree(root);
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/**
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* File: hanota.js
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 移动一个圆盘 */
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function move(src, tar) {
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// 从 src 顶部拿出一个圆盘
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const pan = src.pop();
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// 将圆盘放入 tar 顶部
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tar.push(pan);
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}
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/* 求解汉诺塔：问题 f(i) */
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function dfs(i, src, buf, tar) {
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// 若 src 只剩下一个圆盘，则直接将其移到 tar
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if (i === 1) {
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move(src, tar);
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return;
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}
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// 子问题 f(i-1) ：将 src 顶部 i-1 个圆盘借助 tar 移到 buf
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dfs(i - 1, src, tar, buf);
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// 子问题 f(1) ：将 src 剩余一个圆盘移到 tar
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move(src, tar);
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// 子问题 f(i-1) ：将 buf 顶部 i-1 个圆盘借助 src 移到 tar
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dfs(i - 1, buf, src, tar);
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}
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/* 求解汉诺塔 */
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function solveHanota(A, B, C) {
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const n = A.length;
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// 将 A 顶部 n 个圆盘借助 B 移到 C
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dfs(n, A, B, C);
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}
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/* Driver Code */
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// 列表尾部是柱子顶部
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const A = [5, 4, 3, 2, 1];
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const B = [];
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const C = [];
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console.log('初始状态下：');
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console.log(`A = ${JSON.stringify(A)}`);
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console.log(`B = ${JSON.stringify(B)}`);
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console.log(`C = ${JSON.stringify(C)}`);
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solveHanota(A, B, C);
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console.log('圆盘移动完成后：');
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console.log(`A = ${JSON.stringify(A)}`);
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console.log(`B = ${JSON.stringify(B)}`);
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console.log(`C = ${JSON.stringify(C)}`);
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@ -0,0 +1,54 @@
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/**
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* File: subset_sum_i.ts
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 回溯算法：子集和 I */
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function backtrack(
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state: number[],
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target: number,
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choices: number[],
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start: number,
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res: number[][]
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): void {
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// 子集和等于 target 时，记录解
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if (target === 0) {
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res.push([...state]);
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return;
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}
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// 遍历所有选择
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// 剪枝二：从 start 开始遍历，避免生成重复子集
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for (let i = start; i < choices.length; i++) {
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// 剪枝一：若子集和超过 target ，则直接结束循环
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// 这是因为数组已排序，后边元素更大，子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 尝试：做出选择，更新 target, start
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state.push(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res);
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// 回退：撤销选择，恢复到之前的状态
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state.pop();
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}
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}
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/* 求解子集和 I */
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function subsetSumI(nums: number[], target: number): number[][] {
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const state = []; // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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const start = 0; // 遍历起始点
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const res = []; // 结果列表（子集列表）
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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const nums = [3, 4, 5];
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const target = 9;
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const res = subsetSumI(nums, target);
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console.log(`输入数组 nums = ${JSON.stringify(nums)}, target = ${target}`);
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console.log(`所有和等于 ${target} 的子集 res = ${JSON.stringify(res)}`);
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export {};
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/**
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* File: subset_sum_ii.ts
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 回溯算法：子集和 II */
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function backtrack(
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state: number[],
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target: number,
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choices: number[],
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start: number,
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res: number[][]
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): void {
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// 子集和等于 target 时，记录解
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if (target === 0) {
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res.push([...state]);
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return;
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}
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// 遍历所有选择
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// 剪枝二：从 start 开始遍历，避免生成重复子集
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// 剪枝三：从 start 开始遍历，避免重复选择同一元素
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for (let i = start; i < choices.length; i++) {
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// 剪枝一：若子集和超过 target ，则直接结束循环
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// 这是因为数组已排序，后边元素更大，子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
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if (i > start && choices[i] === choices[i - 1]) {
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continue;
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}
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// 尝试：做出选择，更新 target, start
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state.push(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i + 1, res);
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// 回退：撤销选择，恢复到之前的状态
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state.pop();
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}
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}
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/* 求解子集和 II */
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function subsetSumII(nums: number[], target: number): number[][] {
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const state = []; // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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const start = 0; // 遍历起始点
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const res = []; // 结果列表（子集列表）
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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const nums = [4, 4, 5];
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const target = 9;
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const res = subsetSumII(nums, target);
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console.log(`输入数组 nums = ${JSON.stringify(nums)}, target = ${target}`);
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console.log(`所有和等于 ${target} 的子集 res = ${JSON.stringify(res)}`);
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export {};
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@ -0,0 +1,41 @@
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/**
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* File: binary_search_recur.ts
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
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*/
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/* 二分查找：问题 f(i, j) */
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function dfs(nums: number[], target: number, i: number, j: number): number {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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const m = i + ((j - i) >> 1);
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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function binarySearch(nums: number[], target: number): number {
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const n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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/* Driver Code */
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const target = 6;
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const nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35];
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// 二分查找（双闭区间）
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const index = binarySearch(nums, target);
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console.log(`目标元素 6 的索引 = ${index}`);
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export {};
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@ -0,0 +1,51 @@
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/**
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* File: build_tree.ts
|
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* Created Time: 2023-07-30
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* Author: yuan0221 (yl1452491917@gmail.com)
|
||||
*/
|
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import { printTree } from '../modules/PrintUtil';
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import { TreeNode } from '../modules/TreeNode';
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|
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/* 构建二叉树：分治 */
|
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function dfs(
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preorder: number[],
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inorder: number[],
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hmap: Map<number, number>,
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i: number,
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l: number,
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r: number
|
||||
): TreeNode | null {
|
||||
// 子树区间为空时终止
|
||||
if (r - l < 0) return null;
|
||||
// 初始化根节点
|
||||
const root: TreeNode = new TreeNode(preorder[i]);
|
||||
// 查询 m ，从而划分左右子树
|
||||
const m = hmap.get(preorder[i]);
|
||||
// 子问题：构建左子树
|
||||
root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1);
|
||||
// 子问题：构建右子树
|
||||
root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r);
|
||||
// 返回根节点
|
||||
return root;
|
||||
}
|
||||
|
||||
/* 构建二叉树 */
|
||||
function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
|
||||
// 初始化哈希表，存储 inorder 元素到索引的映射
|
||||
let hmap = new Map<number, number>();
|
||||
for (let i = 0; i < inorder.length; i++) {
|
||||
hmap.set(inorder[i], i);
|
||||
}
|
||||
const root = dfs(preorder, inorder, hmap, 0, 0, inorder.length - 1);
|
||||
return root;
|
||||
}
|
||||
|
||||
/* Driver Code */
|
||||
const preorder = [3, 9, 2, 1, 7];
|
||||
const inorder = [9, 3, 1, 2, 7];
|
||||
console.log('前序遍历 = ' + JSON.stringify(preorder));
|
||||
console.log('中序遍历 = ' + JSON.stringify(inorder));
|
||||
const root = buildTree(preorder, inorder);
|
||||
console.log('构建的二叉树为：');
|
||||
printTree(root);
|
||||
@ -0,0 +1,52 @@
|
||||
/**
|
||||
* File: hanota.ts
|
||||
* Created Time: 2023-07-30
|
||||
* Author: yuan0221 (yl1452491917@gmail.com)
|
||||
*/
|
||||
|
||||
/* 移动一个圆盘 */
|
||||
function move(src: number[], tar: number[]): void {
|
||||
// 从 src 顶部拿出一个圆盘
|
||||
const pan = src.pop();
|
||||
// 将圆盘放入 tar 顶部
|
||||
tar.push(pan);
|
||||
}
|
||||
|
||||
/* 求解汉诺塔：问题 f(i) */
|
||||
function dfs(i: number, src: number[], buf: number[], tar: number[]): void {
|
||||
// 若 src 只剩下一个圆盘，则直接将其移到 tar
|
||||
if (i === 1) {
|
||||
move(src, tar);
|
||||
return;
|
||||
}
|
||||
// 子问题 f(i-1) ：将 src 顶部 i-1 个圆盘借助 tar 移到 buf
|
||||
dfs(i - 1, src, tar, buf);
|
||||
// 子问题 f(1) ：将 src 剩余一个圆盘移到 tar
|
||||
move(src, tar);
|
||||
// 子问题 f(i-1) ：将 buf 顶部 i-1 个圆盘借助 src 移到 tar
|
||||
dfs(i - 1, buf, src, tar);
|
||||
}
|
||||
|
||||
/* 求解汉诺塔 */
|
||||
function solveHanota(A: number[], B: number[], C: number[]): void {
|
||||
const n = A.length;
|
||||
// 将 A 顶部 n 个圆盘借助 B 移到 C
|
||||
dfs(n, A, B, C);
|
||||
}
|
||||
|
||||
/* Driver Code */
|
||||
// 列表尾部是柱子顶部
|
||||
const A = [5, 4, 3, 2, 1];
|
||||
const B = [];
|
||||
const C = [];
|
||||
console.log('初始状态下：');
|
||||
console.log(`A = ${JSON.stringify(A)}`);
|
||||
console.log(`B = ${JSON.stringify(B)}`);
|
||||
console.log(`C = ${JSON.stringify(C)}`);
|
||||
|
||||
solveHanota(A, B, C);
|
||||
|
||||
console.log('圆盘移动完成后：');
|
||||
console.log(`A = ${JSON.stringify(A)}`);
|
||||
console.log(`B = ${JSON.stringify(B)}`);
|
||||
console.log(`C = ${JSON.stringify(C)}`);
|
||||
Loading…
Reference in new issue