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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((a, b) => a - b); // 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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