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){
10
if(item.size()==k){
11 res.add(
new ArrayList<Integer>(item));
//
because item is ArrayList<T> so it will not disappear from stack to stack
12
return;
13 }
14
for(
int i=start;i<=n;i++){
15 item.add(i);
16 dfs(n,k,i+1,item,res);
17 item.remove(item.size()-1);
18 }
19 }
Reference:http://blog.csdn.net/linhuanmars/article/details/21260217