第十五周：[leetCode] 90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

解题思路：寻找数组中不重复的所有子集；可立即反应用深度搜索，通关条件判断除去重复项，dps基本框架为：

void dfs(&nums, start, sub, &res) {    if(condition) return;    else        res.push_back(sub);    for(int i = start; i < nums.size(); i++) {        sub.push_back(nums[i]);        dfs(nums, i + 1, sub, res);        sub.pop_back();    }    return;}

思考通过改变判断条件即可实现输出不重复子集：

void dfs(vector<int>& nums, int start, vector<int> sub, vector<vector<int>>& res){   res.push_back(sub);   for(int i = start; i < nums.size(); i++){       if(i!=start && nums[i] == nums[i-1]) continue;       sub.push_back(nums[i]);       dfs(nums, i+1, sub, res);       sub.pop_back();   }   return;}vector<vector<int>> subsetsWithDup(vector<int>& nums) {   vector<vector<int>> res;   vector<int> sub;   sort(nums.begin(), nums.end());   dfs(nums, 0, sub, res);   return res;}