标签：leetcode
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Ensure that numbers within the set are sorted in ascending order.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
void combinationSum(int k,int index,int n,vector<int>& tmp,vector< vector<int> >& vec)
{//k个数据相加为n，数据范围是index~9,tmp用于临时存放数据，vec用于存放返回值
if(k==1)
{//还有最后一个名额，如果找不到相符的数据，说明没有满足的情况，需要返回
for(int i=index;i<=9;++i)
{
if(i==n)
{
tmp[k-1]=i;
vec.push_back(tmp);
break;
}
}
}
else
{
for(int i=index;i<=9;++i)
{
tmp[k-1]=i;//反向存放数据,递归
combinationSum(k-1,i+1,n-i,tmp,vec);
}
}
}
vector< vector<int> > combinationSum3(int k, int n)
{
vector< vector<int> > vec;
vector<int> tmp;
for(int i=0;i<k;++i)
{//初始化
tmp.push_back(0);
}
combinationSum(k,1,n,tmp,vec);
for(int i=0;i<vec.size();++i)
{//因为数据是反向存放的，所以还需要一次反转
reverse(vec[i].begin(),vec[i].end());
}
return vec;
}
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leetCode(31):Combination Sum III
标签：leetcode
原文地址：http://blog.csdn.net/walker19900515/article/details/46785831
