Add kotlin code for the chapter of divide and conquer (#1098)
* feat(kotlin):new kotlin support files * fix(kotlin): reviewed the formatting, comments and so on. * fix(kotlin): fix the indentation and format * feat(kotlin): Add kotlin code for the backtraking chapter. * fix(kotlin): fix incorrect output of preorder_traversal_iii_template.kt file * fix(kotlin): simplify kotlin codes * fix(kotlin): modify n_queens.kt for consistency. * feat(kotlin): add kotlin code for computational complexity. * fix(kotlin): remove iteration folder. * fix(kotlin): remove n_queens.kt file out of folder. * fix(kotlin): remove some folders. * style(kotlin): modified two chapters. * feat(kotlin): add kotlin code for divide and conquer. * Update build_tree.kt * Update hanota.kt * Delete codes/kotlin/chapter_backtracking directory * Delete codes/kotlin/chapter_computational_complexity directory * Delete codes/kotlin/chapter_divide_and_conquer directory * feat(kotlin): add kotlin code for divide and conquer. * Update hanota.ktpull/1099/head
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/**
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* File: binary_search_recur.kt
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* Created Time: 2024-01-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_divide_and_conquer.binary_search_recur
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/* 二分查找:问题 f(i, j) */
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fun dfs(
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nums: IntArray,
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target: Int,
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i: Int,
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j: Int
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): Int {
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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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val m = (i + j) / 2
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return if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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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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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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m
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}
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}
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/* 二分查找 */
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fun binarySearch(nums: IntArray, target: Int): Int {
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val n = nums.size
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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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fun main() {
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val target = 6
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val nums = intArrayOf(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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val index = binarySearch(nums, target)
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println("目标元素 6 的索引 = $index")
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}
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/**
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* File: build_tree.kt
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* Created Time: 2024-01-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_divide_and_conquer.build_tree
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import utils.TreeNode
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import utils.printTree
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/* 构建二叉树:分治 */
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fun dfs(preorder: IntArray, inorderMap: Map<Int?, Int?>, i: Int, l: Int, r: Int): TreeNode? {
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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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val root = TreeNode(preorder[i])
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// 查询 m ,从而划分左右子树
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val m = inorderMap[preorder[i]]!!
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// 子问题:构建左子树
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root.left = dfs(preorder, inorderMap, 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, inorderMap, 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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fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? {
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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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val inorderMap: MutableMap<Int?, Int?> = HashMap()
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for (i in inorder.indices) {
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inorderMap[inorder[i]] = i
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}
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val root = dfs(preorder, inorderMap, 0, 0, inorder.size - 1)
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return root
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}
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/* Driver Code */
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fun main() {
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val preorder = intArrayOf(3, 9, 2, 1, 7)
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val inorder = intArrayOf(9, 3, 1, 2, 7)
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println("前序遍历 = " + preorder.contentToString())
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println("中序遍历 = " + inorder.contentToString())
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val root = buildTree(preorder, inorder)
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println("构建的二叉树为:")
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printTree(root)
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}
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/**
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* File: hanota.kt
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* Created Time: 2024-1-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_divide_and_conquer.hanota
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/* 移动一个圆盘 */
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fun move(src: MutableList<Int>, tar: MutableList<Int>) {
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// 从 src 顶部拿出一个圆盘
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val pan: Int = src.removeAt(src.size - 1)
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// 将圆盘放入 tar 顶部
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tar.add(pan)
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}
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/* 求解汉诺塔问题 f(i) */
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fun dfs(i: Int, src: MutableList<Int>, buf: MutableList<Int>, tar: MutableList<Int>) {
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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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/* 求解汉诺塔问题 */
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fun solveHanota(A: MutableList<Int>, B: MutableList<Int>, C: MutableList<Int>) {
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val n = A.size
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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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fun main() {
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// 列表尾部是柱子顶部
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val A: MutableList<Int> = ArrayList(mutableListOf(5, 4, 3, 2, 1))
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val B: MutableList<Int> = ArrayList()
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val C: MutableList<Int> = ArrayList()
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println("初始状态下:")
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println("A = $A")
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println("B = $B")
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println("C = $C")
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solveHanota(A, B, C)
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println("圆盘移动完成后:")
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println("A = $A")
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println("B = $B")
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println("C = $C")
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}
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