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, … , 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：解题思路跟Combination Sum类似，但在同一层递归时要跳过相同的元素。代码如下：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > result;
        vector<int> path;
        
        sort(num.begin(), num.end());
        dfs(result, path, num, 0, target);
        
        return result;
    }
    
    void dfs(vector<vector<int> > &result, vector<int> &path, vector<int> &num, int start, int gap){
        if(gap == 0){
            result.push_back(path);
            return;
        }
        
        for(int i = start; i < num.size(); i++){
            if(gap < num[i]) return; // prune
            
            if(i == start || num[i-1] != num[i]){
                path.push_back(num[i]);
                dfs(result, path, num, i+1, gap-num[i]);
                path.pop_back();
            }
        }
    }
};