[LeetCode]040-Combination Sum II
2015-12-29 14:45
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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:
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]
Solution:
思路,与上一题类似,不过在移位时,遍历时往后移一位,同时申请一个map用于除重。
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]
Solution:
思路,与上一题类似,不过在移位时,遍历时往后移一位,同时申请一个map用于除重。
class Solution { public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<int> result; vector<vector<int>> results; map<vector<int>,int> checkmap; sort(candidates.begin(),candidates.end()); combine(candidates,0,result,results,target,checkmap); return results; } void combine(vector<int>& candidates,int begin,vector<int>& result,vector<vector<int>>& results,int target,map<vector<int>,int>&checkmap) { int n = candidates.size(); int i = 0; for(i = begin;i<n;i++) { if(candidates[i] == target) { result.push_back(candidates[i]); if(checkmap.find(result) == checkmap.end()) { results.push_back(result); checkmap.insert(pair<vector<int>,int>(result,i)); } result.pop_back(); return; } else if(candidates[i] < target) { int t = target - candidates[i]; result.push_back(candidates[i]); combine(candidates,i+1,result,results,t,checkmap); if(!result.empty()) { result.pop_back(); } } else return; } } };
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