#Leetcode# 40. Combination Sum II

https://leetcode.com/problems/combination-sum-ii/

 

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

代码：

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        int n = candidates.size();
        sort(candidates.begin(), candidates.end());
        set<vector<int> > ans;
        vector<int> out;
        vector<int> vis(n, 0);
        
        dfs(candidates, out, ans, target, 0, vis);
        return vector<vector<int>>(ans.begin(), ans.end());
        
    }
    void dfs(vector<int> &candidates, vector<int> &out, set<vector<int> > &ans, int key, int st, vector<int> &vis) {
        if(key < 0) return ;
        else if(key == 0) {
            ans.insert(out);
            return;
        }
        for(int i = st; i < candidates.size(); i ++) {
            if(vis[i] == 0) {
                vis[i] = 1;
                out.push_back(candidates[i]);
                dfs(candidates, out, ans, key - candidates[i], i, vis);
                vis[i] = 0;
                out.pop_back();
            }
        }
    } 
};