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],
]

题解:

这道题就是用DFS(参考Work BreakII)的循环递归处理子问题的方法解决。n为循环的次数,k为每次尝试的不同数字。用到了回溯。

代码如下:

 1     public ArrayList<ArrayList<Integer>> combine(int n, int k) {
 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
 3         if(n <= 0||n < k)
 4             return res;
 5         ArrayList<Integer> item = new ArrayList<Integer>();    
 6         dfs(n,k,1,item, res);//because it need to begin from 1
 7         return res;
 8     }
 9     private void dfs(int n, int k, int start, ArrayList<Integer> item, ArrayList<ArrayList<Integer>> res){
         if(item.size()==k){
             res.add(new ArrayList<Integer>(item));//because item is ArrayList<T> so it will not disappear from stack to stack
             return;
         }
         for(int i=start;i<=n;i++){
             item.add(i);
             dfs(n,k,i+1,item,res);
             item.remove(item.size()-1);
         }
     }

Reference:http://blog.csdn.net/linhuanmars/article/details/21260217

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