题目：子集

思路

回溯法

代码

class Solution {
public:vector<vector<int>> result;vector<int> path;void function(vector<int>& nums, int startindex){// 为什么要到这里写? 后面调用递归之前就不对result.push_back(path);// 到头了if(startindex == nums.size()){return;}int i;int size;size = nums.size();for(i = startindex; i < size; i++){path.push_back(nums[i]);function(nums, i+1);path.pop_back();}return;}vector<vector<int>> subsets(vector<int>& nums) {result.clear();path.clear();function(nums, 0);return result;}
};

在回溯函数开头就记录path结果；

错误示范

class Solution {
public:vector<vector<int>> result;vector<int> path;void function(vector<int>& nums, int startindex){// 为什么要到这里写? 后面调用递归之前就不对// 到头了if(startindex == nums.size()){return;}int i;int size;size = nums.size();for(i = startindex; i < size; i++){path.push_back(nums[i]);result.push_back(path);function(nums, i+1);path.pop_back();}return;}vector<vector<int>> subsets(vector<int>& nums) {result.clear();path.clear();function(nums, 0);return result;}
};

在后面调用递归之前写result.push_back(path);就会漏掉空集，因为它进不去for循环；