学步园

题目：

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] 

解题：

基本等同于Combination Sum，dfs下一层时候不是从当前坐标开始，而是从下一个开始。记录答案的时候判一下重。

代码：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        ans.clear();
        single.clear();
        sort(candidates.begin(), candidates.end());
        in = candidates;
        dfs(0, target);
        return ans;
    }
private:
    vector<vector<int> > ans;
    vector<int> single, in;

    void dfs(int pos, int target) {
        if(target == 0) {
            bool flag = 0;
            for(int i = 0; i < ans.size(); i ++) {
                if(ans[i] == single) {
                    flag = 1;
                    break;
                }
            }
            if(!flag)
                ans.push_back(single);
            return ;
        }
        for(int i = pos; i < in.size(); i ++) {
            if(in[i] > target) break;
            single.push_back(in[i]);
            dfs(i + 1, target - in[i]);
            single.pop_back();
        }
        return;
    }
};