Given a collection of numbers, return all possible permutations.
For example,
[1,2,3]
have the following permutations:
[1,2,3]
, [1,3,2]
, [2,1,3]
, [2,3,1]
, [3,1,2]
,
and [3,2,1]
.
给定一组数字,返回所有的可能的组合。
比如:
[1,2,3] 有下面的组合:
[1,2,3]
, [1,3,2]
, [2,1,3]
, [2,3,1]
, [3,1,2]
,
and [3,2,1]
.
建立树,并进行递归 参考博客:http://blog.csdn.net/tuantuanls/article/details/8717262 思路二
建立一棵树,比如说:
public class Solution { public ArrayList<ArrayList<Integer>> permute(int[] num) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); permute(num, 0, result); return result; } void permute(int[] num, int start, ArrayList<ArrayList<Integer>> result) { if (start >= num.length) //终止条件,递归到末尾节点是,将数组转化为链表 { ArrayList<Integer> item = convertArrayToList(num); result.add(item); } for (int j = start; j <= num.length - 1; j++) { swap(num, start, j);//交换 permute(num, start + 1, result);//交换后子代递归 swap(num, start, j);//恢复到交换前的初始状态,以便于得出下一次的交换结果 } } private ArrayList<Integer> convertArrayToList(int[] num) //数组变链表 { ArrayList<Integer> item = new ArrayList<Integer>(); for (int h = 0; h < num.length; h++) item.add(num[h]); return item; } private void swap(int[] a, int i, int j) //交换 { int temp = a[i]; a[i] = a[j]; a[j] = temp; } }
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原文:http://blog.csdn.net/evan123mg/article/details/46875467