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个数字的全部的组合。

例如说。

假设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<ArrayList<Integer>> combine(int n, int k)
{
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(n<=0 || n<k)
return res;
helper(n,k,1,new ArrayList<Integer>(), res);
return res;
}
private void helper(int n, int k, int start, ArrayList<Integer> item, ArrayList<ArrayList<Integer>> res)
{
if(item.size()==k)
{
res.add(new ArrayList<Integer>(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
}
}
}

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