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);

        }

    }

};