LeetCode 040 Combination Sum II

题目要求：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] 

分析：

Combination Sum 里面的元素可以无限次使用，但是Combination Sum II每个元素只能使用一次。

代码如下：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(candidates.begin(), candidates.end());
        set<vector<int> > ans;
        vector<int> record;
        searchAns(ans, record, candidates, target, 0);
        
        vector<vector<int> > temp;
        for (set<vector<int> >::iterator it = ans.begin(); it != ans.end(); it++) {
            temp.push_back(*it);
        }
        return temp;
    }
    
private:
    void searchAns(set<vector<int> > &ans, vector<int> &record, vector<int> &candidates, int target, int idx) {
        
        if (target == 0) {
            ans.insert(record);
            return;
        }
        
        if (idx == candidates.size() || candidates[idx] > target) {
            return;
        }
        
        for (int i = 1; i >= 0; i--) {
            record.push_back(candidates[idx]);            
        }
        
        for (int i = 1; i >= 0; i--) {
            record.pop_back();
            searchAns(ans, record, candidates, target - i * candidates[idx], idx + 1);
        }
    }
};