Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3→1,3,23,2,1→1,2,31,1,5→1,5,1
算法思路：
不得不说，我第一遍做的时候，这道题的算法还是蛮好的，一次遍历，空间O(1)，时间O(nlogn)主要浪费在排序上。
代码如下：
1 public class Solution { 2 public void nextPermutation(int[] num) { 3 if(num == null || num.length <= 1) return ; 4 for(int i = num.length - 1; i > 0; i--){ 5 if(num[i] > num[i - 1]){ 6 int one = i; 7 int two = i - 1; 8 int min = num[i]; 9 for(int j = i ; j < num.length; j++){ 10 if(num[j] > num[i - 1] && num[j] < min){ 11 min = num[j]; 12 one = j; 13 } 14 } 15 num[one] = num[two] + num[one]; 16 num[two] = num[one] - num[two]; 17 num[one] = num[one] - num[two]; 18 Arrays.sort(num, i, num.length); 19 return; 20 } 21 } 22 Arrays.sort(num); 23 return; 24 } 25 }
[leetcode]Next Permutation,布布扣,bubuko.com
原文：http://www.cnblogs.com/huntfor/p/3863730.html
