标签：input 解决 pop can tin find 元素 ack including
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
target) will be positive integers.Example 1:
Input: candidates =[10,1,2,7,6,1,5], target =8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
比起上个题多了个条件,结果不能重复,并且给定的元素有重复;
解决办法,在dfs的每层循环添加元素时,判断和前一个是不是相等,相等则不添加;
class Solution { public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { sort(candidates.begin(),candidates.end()); vector<vector<int>> res; vector<int> tmp; dfs(res,tmp,target,candidates,0); return res; } void dfs(vector<vector<int>> &res, vector<int> &tmp, int target, vector<int> &candidates, int index) { if(0==target) { res.push_back(tmp); return; } for(int i=index;i<candidates.size()&&candidates[i]<=target;++i) //注意加上 candidates<=targe, 否则容易超时 { if(i!=index&&candidates[i]==candidates[i-1])continue; tmp.push_back(candidates[i]); dfs(res,tmp,target-candidates[i], candidates,i+1); tmp.pop_back(); } } };
标签：input 解决 pop can tin find 元素 ack including
原文地址：https://www.cnblogs.com/lychnis/p/11704893.html
