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] ]
题目大意:
从给定的数组中挑选元素，使得这些元素的和为target。元素不能重复。和题目39. Combination Sum 的唯一差别就是相同元素只能使用一次。
题目代码:
class Solution {
public:vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {sort(candidates.begin(), candidates.end());dfs(0, 0, candidates, target);return ans;}
private:vector<vector<int>> ans;vector<int> elements;void dfs(int cur, int sum, vector<int>& candidates, int target){if(sum == target){if(!count(ans.begin(), ans.end(), elements))ans.push_back(elements);return ;}if(sum > target) return ;for(int i = cur;i < candidates.size(); i++){elements.push_back(candidates[i]);dfs(i+1, sum+candidates[i], candidates, target);elements.pop_back();}}
};