[LeetCode-Algorithms-40] "Combination Sum II" (2017.11.29-WEEK13)

题目链接：Combination Sum II

• 题目描述：
  

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]]

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(1)思路：递归回溯，但注意在同一层递归树中，如果某元素已经处理并进入下一层递归，那么与该元素相同的值就应该跳过,否则将出现重复;另外，相同元素第一个进入下一层递归，而不是任意一个。

(2)代码：

class Solution {public:    vector<vector<int> > combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int> > ret;        vector<int> cur;        Helper(ret, cur, candidates, target, 0);        return ret;    }    void Helper(vector<vector<int> > &ret, vector<int> cur, vector<int> &candidates, int target, int position){        if(target == 0)            ret.push_back(cur);        else{            for(int i = position; i < candidates.size() &&candidates[i] <= target; i ++){                if(i != position && candidates[i] == candidates[i-1])                    continue;                cur.push_back(candidates[i]);                Helper(ret, cur, candidates, target-candidates[i], i+1);                cur.pop_back();            }        }    }};

(3)提交结果：