[LeetCode]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.
• 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] 

解题思路：

similar to (http://www.cnblogs.com/skycore/p/5261412.html)

 1 class Solution {
 2 public:
 3     vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
 4         sort(candidates.begin(), candidates.end());
 5         vector<vector<int>> result;
 6         vector<int> elem;
 7         combinationSum2(candidates, target, 0, result, elem);
 8         return result;
 9     }
10 private:
11     void combinationSum2(vector<int>& candidates, int target, int index, vector<vector<int>> &result, vector<int> &elem) {
12         if (target == 0) {
13             result.push_back(elem);
14             return;
15         }
16         
17         for (int i = index; i < candidates.size() && candidates[i] <= target; ++i) {
18             if (i > index && candidates[i] == candidates[i - 1]) continue;
19             elem.push_back(candidates[i]);
20             combinationSum2(candidates, target - candidates[i], i + 1, result, elem);
21             elem.pop_back();
22         }
23     }
24 };

 

 

 

 

转载于:https://www.cnblogs.com/skycore/p/5261484.html