LeetCode讲解篇之40. 组合总和 II

题目描述

题解思路

按升序排序candidates，然后遍历candidates，目标数减去当前candidates的数，若该结果小于0，因为candidates的元素大于0，所以后续不会再出现让计算结果等于0的情况，所以直接break，如果该结果等于0，将数据加到结果集合中，然后break，若该结果大于0，则将当前candidates的数加入tmp数组，递归调用，调用结束后，删去tmp中添加的当前candidates的数，然后去掉和当前candidates的数重复的数

题解代码

func combinationSum2(candidates []int, target int) [][]int {sort.Ints(candidates)res := make([][]int, 0)var dfs func([]int, int)tmp := []int{}dfs = func(candidates []int, target int) {n := len(candidates)for i := 0; i < n; i++ {temp := target - candidates[i]if temp < 0 {break}if temp == 0 {res = append(res, append([]int{}, append(tmp, candidates[i])...))break}tmp = append(tmp, candidates[i])dfs(candidates[i+1:], temp)tmp = tmp[:len(tmp)-1]for i < n - 1 && candidates[i] == candidates[i+1] {i++}}}dfs(candidates, target)return res
}