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.

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：

 1 class Solution {
 2 private:
 3     vector<vector<int>>result;
 4     void dfs(int start, int end, const vector<int>& candidates, int target, vector<int>& internal) {
 5         if (start > end) {
 6             return;
 7         }
 8         if (candidates[start] == target) {
 9             internal.push_back(candidates[start]);
10             if ( find(result.begin(), result.end(), internal) == result.end()) {
11                 result.push_back(internal);
12             }
13             internal.pop_back();
14             return;
15         }
16         if (candidates[start] > target) {
17             return;
18         }
19         dfs(start + 1, end, candidates, target, internal);
20         internal.push_back(candidates[start]);
21         dfs(start + 1, end, candidates, target - candidates[start], internal);
22
23         internal.pop_back();
24     }
25 public:
26     vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
27         sort(candidates.begin(), candidates.end());
28         int end = candidates.size() - 1;
29         vector<int> internal;
30         dfs(0, end, candidates, target, internal);
31         return result;
32     }
33 };