题目：
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example, If n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
题意:
给定两个整数 n 和 k，返回1 ...n中k个数字的所有的组合。
比方说，
如果n=4 and k=2,解为：
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
算法分析：
用一个循环递归处理子问题。
AC代码：
public class Solution
{
public ArrayList> combine(int n, int k)
{
ArrayList> res = new ArrayList>();
if(n<=0 || n
return res;
helper(n,k,1,new ArrayList(), res);
return res;
}
private void helper(int n, int k, int start, ArrayList item, ArrayList> res)
{
if(item.size()==k)
{
res.add(new ArrayList(item));
return;
}
for(int i=start;i<=n;i++) // try each possibility number in current position
{
item.add(i);
helper(n,k,i+1,item,res); // after selecting number for current position, process next position
item.remove(item.size()-1); // clear the current position to try next possible number
}
}
}
