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leetcode之Combination Sum

2014-08-22 15:17:00.0 leetcode 算法 Combination Sum 递归 leetcode  
导读:Combination SumGiven a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.The same repeated number may be chosen from C unlim...。。。

Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7
A solution set is: 
[7] 
[2, 2, 3] 

思路:典型的递归程序,对每次遍历,取当前数和不取当前数进行遍历。

class Solution {public:    void combinationSum(vector& candidates,int index,int currentSum,int target,vector& path)    {     if(currentSum == target)     {      res.push_back(path);      return;     }     int length = candidates.size();     if(index == length || currentSum > target)return;     path.push_back(candidates[index]);     combinationSum(candidates,index,currentSum+candidates[index],target,path);//重复添加当前节点     path.pop_back();     combinationSum(candidates,index+1,currentSum,target,path);//跳过当前节点    }        vector > combinationSum(vector &candidates, int target) {        sort(candidates.begin(),candidates.end());     vector path;     combinationSum(candidates,0,0,target,path);     return res;    }private:    vector > res;};

Combination Sum II

 

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 (a1a2, … , 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] 

class Solution {public:    void combinationSum2(vector& num,int index,int currentSum,int target,vector& path)    {     if(currentSum == target)     {      hash.insert(path);//防止重复      return;     }     int length = num.size();     if(index == length || currentSum > target)return;     path.push_back(num[index]);     combinationSum2(num,index+1,currentSum+num[index],target,path);//取当前节点     path.pop_back();     combinationSum2(num,index+1,currentSum,target,path);//不取当前节点    }        vector > combinationSum2(vector &num, int target) {     sort(num.begin(),num.end());     vector path;     combinationSum2(num,0,0,target,path);     vector > res(hash.begin(),hash.end());     return res;    }private:    set > hash;};

如果是求总数,而且不需要去除重复,可以看这里



(编辑: fangjian1204)

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