leetCode(31):Combination Sum III 分类： leetCode 2015-07-07 10:23 163人阅读 评论(0) 收藏

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]]

基本思路：首先确定第一个数，然后在剩余的数列中去寻找其余的数，找到则满足；说到这，自然而然--》递归

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;
}