leetcode:子集（dfs+位运算两种解法）

思路&＃xff1a;
1.主体思路跟“组合”一致
2.退出条件改为tempindex超出
3.只要长度小于等于len都可以加入res

代码&＃xff1a;

func subsets(nums []int) [][]int {res &＃61; [][]int{}numsLen :&＃61; len(nums)find(0, []int{}, nums, numsLen)return res
}func find(tempindex int, tempres, nums []int, k int) {fmt.Println(tempres)if tempindex > k {return}if len(tempres) <&＃61; k {comb :&＃61; make([]int, len(tempres))copy(comb, tempres)res &＃61; append(res, comb)}for i :&＃61; tempindex; i < k; i&＃43;&＃43; {tempres &＃61; append(tempres, nums[i])find(i &＃43; 1, tempres, nums, k)tempres &＃61; tempres[:len(tempres) - 1]}
}


位运算代表01&＃xff1a;

func subsets(nums []int) (ans [][]int) {n :&＃61; len(nums)//从0到1<for mask :&＃61; 0; mask < 1<<n; mask&＃43;&＃43; {set :&＃61; []int{}for i, v :&＃61; range nums {//看看mask指示的第i位&＃xff08;从右到左&＃xff09;是否取1//是的话加入这个index的valueif mask>>i&1 > 0 {set &＃61; append(set, v)}}ans &＃61; append(ans, append([]int(nil), set...))}return
}