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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
Solution:
http://fisherlei.blogspot.com/2012/12/leetcode-next-permutation.html
算法思想如图所示：
代码如下：
1 public class Solution { 2 public void nextPermutation(int[] num) { 3 if(num.length<2) 4 return; 5 int N=num.length; 6 int index=N-1; 7 while(index>0){ 8 if(num[index]<=num[index-1]) 9 index--; 10 else 11 break; 12 } 13 if(index==0){ 14 Arrays.sort(num); 15 return; 16 } 17 18 int val=num[index-1]; 19 //System.out.println(val); 20 int j=N-1; 21 while(j>index-1){ 22 if(num[j]>val){ 23 break; 24 } 25 else 26 j--; 27 } 28 // System.out.println(num[j]); 29 int temp=val; 30 num[index-1]=num[j]; 31 num[j]=temp; 32 Arrays.sort(num, index, N); 33 } 34 }
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原文地址：http://www.cnblogs.com/Phoebe815/p/4194946.html
