题目地址
https://leetcode.com/problems/combination-sum/description/
题目描述
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
思路
这道题目是求集合，并不是求极值，因此动态规划不是特别切合，因此我们需要考虑别的方法。
这种题目其实有一个通用的解法，就是回溯法。 网上也有大神给出了这种回溯法解题的 通用写法，这里的所有的解法使用通用方法解答。 除了这道题目还有很多其他题目可以用这种通用解法，具体的题目见后方相关题目部分。
我们先来看下通用解法的解题思路，我画了一张图：
通用写法的具体代码见下方代码区。
关键点解析
- 回溯法
- backtrack 解题公式
代码
/*
* @lc app=leetcode id=39 lang=javascript
*
* [39] Combination Sum
*
* https://leetcode.com/problems/combination-sum/description/
*
* algorithms
* Medium (46.89%)
* Total Accepted: 326.7K
* Total Submissions: 684.2K
* Testcase Example: '[2,3,6,7]\n7'
*
* Given a set of candidate numbers (candidates) (without duplicates) and a
* target number (target), find all unique combinations in candidates where the
* candidate numbers sums to target.
*
* The same repeated number may be chosen from candidates unlimited number of
* times.
*
* Note:
*
*
* All numbers (including target) will be positive integers.
* The solution set must not contain duplicate combinations.
*
*
* Example 1:
*
*
* Input: candidates = [2,3,6,7], target = 7,
* A solution set is:
* [
* [7],
* [2,2,3]
* ]
*
*
* Example 2:
*
*
* Input: candidates = [2,3,5], target = 8,
* A solution set is:
* [
* [2,2,2,2],
* [2,3,3],
* [3,5]
* ]
*
*/
function backtrack(list, tempList, nums, remain, start) {
if (remain < 0) return;
else if (remain === 0) return list.push([...tempList]);
for (let i = start; i < nums.length; i++) {
tempList.push(nums[i]);
backtrack(list, tempList, nums, remain - nums[i], i); // 数字可以重复使用， i + 1代表不可以重复利用
tempList.pop();
}
}
/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
var combinationSum = function(candidates, target) {
const list = [];
backtrack(list, [], candidates.sort((a, b) => a - b), target, 0);
return list;
};
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