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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 };
LeetCode40 Combination Sum II
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