【LeetCode】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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• 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] 

 

思路：

同样DFS，跟上一题是一样的，只不过为了不重复，每次调用dfs的时候pos = i + 1；

这道题备选数组是有重复的，得去重

 

代码：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<int> temp;
        vector<vector<int> > rst;
        sort(num.begin(), num.end());
        dfs(num, target, temp, rst, 0);
        return rst;
    }

    void dfs(vector<int> &num, int target, vector<int> &temp, vector<vector<int> > &rst, int pos)
    {
        if (target < 0)
            return;
        if (target == 0)
        {
            rst.push_back(temp);
            return;
        }

        for (int i = pos; i < num.size(); i++)
        {
            temp.push_back(num[i]);
            dfs(num, target - num[i], temp, rst, i + 1);
            temp.pop_back();
            while (i < num.size() - 1 && num[i] == num[i+1])
                i++;
        }
    }
};