LeetCode OJ Combination Sum II

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

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

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 

[1, 1, 6] 

class Solution {public:    vector<vector<int> > ans;  // answer    vector<int> can;  // "can" is a vector from candidates after processing    vector<bool> used;    int t;    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        ans.clear();        can = num;        sort(can.begin(), can.end());  // to make sure that the solutions in ans is in non-descending order        used.resize(can.size());        t = target;        dfs(0, 0);        sort(ans.begin(), ans.end());        vector<vector<int> > trueAns;        for (int i = 0; i < ans.size(); i++) {            if (trueAns.size() > 0 && trueAns.back() == ans[i]) continue;            trueAns.push_back(ans[i]);        }        return trueAns;    }    void dfs(int pos, int nowSum) {        if (nowSum == t) {            vector<int> anAns;            for (int i = 0; i < can.size(); i++) {                if (used[i]) anAns.push_back(can[i]);            }            ans.push_back(anAns);            return;        }        if (nowSum > t || pos >= can.size()) return;        used[pos] = false;        dfs(pos + 1, nowSum);        if (can[pos] + nowSum <= t) {            used[pos] = true;            dfs(pos + 1, can[pos] + nowSum);            used[pos] = false;        }    }};