40. Combination Sum II
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题目:
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in Cwhere 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]
给定一个候选数字集合C和目标数字T,找到集合C中所有数字组合和为T的组合,数字集合C中的数字都只允许被使用一次。
note:
1、所有的数字(包括目标数字)都是正整数;
2、所有组合中的数字必须是递增的序列
3、结果集中必须不包含相同的组合
思路:
采用DFS深搜,因为组合的个数不一,需要动态的去判断是否需要加入组合中,是否需要将组合加入结果集中,DFS适合这样的操作。
代码:16ms
class Solution {public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { sort(candidates.begin(), candidates.end()); vector<vector<int>> result; vector<int> cur; combinationSum2(candidates, result, cur, target, 0); return result; } void combinationSum2(vector<int>& candidates, vector<vector<int>>& result, vector<int>& cur, int target, int start){ if(target==0){ result.push_back(cur); return; } if(target<0){ return; } for(int i=start; i<candidates.size(); i++){ if(i>start && candidates[i]==candidates[i-1]){ continue; } cur.push_back(candidates[i]); combinationSum2(candidates, result, cur, target-candidates[i], i+1); cur.pop_back(); } }};
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- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
- 40. Combination Sum II
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