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],
]
分析:
看到返回所有,想到回溯,那么就是DFS了。
public class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> results = new ArrayList<List<Integer>>();
if(k<0 || k>n) return results;
List<Integer> result = new ArrayList<Integer>();
dfs(results, result, n, k, 1);
return results;
}
public void dfs(List<List<Integer>> results, List<Integer> result, int n, int k, int start){
if(result.size()==k){
results.add(new ArrayList<Integer>(result));
return;
}
for(int i=start; i<=n; i++){
//加入结果
result.add(i);
//往后搜索
dfs(results, result, n, k, i+1);
//恢复现场
result.remove(result.size()-1);
}
}
}