递归与回溯搜索

给你一组不重复的整数&＃xff0c;输出几个所有可能的组合顺序&＃xff0c;输出结果不能有重复。

解法一&＃xff1a;递归&＃xff08;空间复杂度太大了&＃xff0c;这个方法不好&＃xff09;

#include "stdafx.h"
#include
#include
#include
using namespace std;void generate(vector& nums, int start, vector>& res, vector& res_sub);
//预测有2的len次幂个结果
vector> subsets(vector& nums) {vector> res;vector res_sub;//插入空集res.push_back(res_sub);generate(nums, 0, res, res_sub);return res;
}
void generate(vector& nums, int start, vector>& res, vector& res_sub) {if (start >&＃61; nums.size()) {return;}//思想&＃xff1a;从左至右依次组合&＃xff0c;记录新组合之后&＃xff0c;去掉队尾&＃xff0c;加入新元素继续向后组合for (int i &＃61; start; i }int main()
{vector nums &＃61; { 1,2,3 };subsets(nums);return 0;
}

解法二&＃xff1a;