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] 

题意：还是和上一题一样，求组合数，但是这里有一个条件是：每个数只能使用一次，不能重复结果集

思路：稍微改一下就行了，一个是每次的下标都+1，另外就是避免重复，跳过相同的数

class Solution {public:    void dfs(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int> &path) {        if (sum > target) return;        if (sum == target) {            res.push_back(path);            return;        }        for (int i = index; i < candidates.size(); i++) {            path.push_back(candidates[i]);            dfs(candidates, i+1, sum+candidates[i], target, res, path);            path.pop_back();            while (i < candidates.size() -1 && candidates[i] == candidates[i+1]) i++;        }    }    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int> > res;        vector<int> path;        dfs(candidates, 0, 0, target, res, path);        return res;    }};