Leetcode 40. Combination Sum II

40. Combination Sum II

• Total Accepted: 76278
• Total Submissions: 265535
• Difficulty: Medium

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.
• 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：还是用回溯法找解，和Leetcode 39. Combination Sum类似。但要注意重复元素造成的干扰，本题排序后，如果有candidates[i]==candidates[i+1]==...==candidates[i+k]，只要对candidates[i]进行回溯求解就可以，剩下的candidates[i+1]...candidates[i+k]直接跳过，所以本层的下一个遍历从candidates[i+k+1]开始。

代码：

class Solution {
public:
    vector<vector<int> > res;
    void combinationSum2(vector<int>& v,vector<int>& candidates,int target,int cur){
        if(!target){
            res.push_back(v);
            return;
        }
        int n=candidates.size(),i;
        for(i=cur;i<n;i++){
            target-=candidates[i];
            if(target>=0){
                v.push_back(candidates[i]);
                combinationSum2(v,candidates,target,i+1);
                v.pop_back();
            }
            else{
                return;
            }
            target+=candidates[i];
            while(i+1<n&&candidates[i]==candidates[i+1]) i++;//去除重复元素导致的重复场景
        }
    }
    vector<vector<int> > combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int> v;
        combinationSum2(v,candidates,target,0);
        return res;
    }
};