LeetCode40 Combination Sum II

题目：

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. (Medium)

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]
]

 分析：

主体回溯框架和Combination Sum I一致，不同之处在于一个元素不能重复用，同时不能出现重复的组合。

所以DFS那一个条件 start 变为start + 1；

同时插入result之前判定是否出现过重复的结果。

代码：

class Solution {
private:
    vector<vector<int>>result;
    void dfs(int start, int end, const vector<int>& candidates, int target, vector<int>& internal) {
        if (start > end) {
            return;
        }
        if (candidates[start] == target) {
            internal.push_back(candidates[start]);
            if ( find(result.begin(), result.end(), internal) == result.end()) {
                result.push_back(internal);
            }
            internal.pop_back();
            return;   
        }
        if (candidates[start] > target) {
            return;
        }
        dfs(start + 1, end, candidates, target, internal);
        internal.push_back(candidates[start]);
        dfs(start + 1, end, candidates, target - candidates[start], internal);

        internal.pop_back();
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        int end = candidates.size() - 1;
        vector<int> internal;
        dfs(0, end, candidates, target, internal);
        return result;
    }
};