leetcode 31. Next Permutation

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 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

找到一串数列的下一个排列.要求不分配额外的内存.

观察数的全排列,发现一个规律.

例如,[5,9,8,4,7,6,2,1],其下一个排列相当于[4,7,6,2,1]的下一个排列,即最后一个小于后面数的数.

然后找到比4大的最小的数6,和4交换[6,7,4,2,1].

然后从第二个数开始逆序排列[6,1,2,4,7].

[5,9,8,6,1,2,4,7],完成.

public class A31NextPermutation {public void nextPermutation(int[] nums) {if(nums != null && nums.length > 1) {int i, j, make;// 找到要交换的第一个数for(i = nums.length - 1; i >= 1 && nums[i] <= nums[i - 1]; i--);if(i != 0) {make = nums[i - 1];// 找到要交换的第二个数for(j = nums.length - 1; j > i - 1 && nums[j] <= make; j--);// 交换nums[i - 1] = nums[j];nums[j] = make;}// 逆序排列(首尾交换)j = nums.length - 1;while(i <= j) {make = nums[i];nums[i] = nums[j];nums[j] = make;i++;j--;}}    }}