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

一刷ac
解题思路：从后向前找，找到一个前一个小于后一个的一组数，然后再后面找比前一个大的最小的那个值，然后两个换位，然后将前一个后面的数从小到大排序。

O(n)

public class Solution {    public void nextPermutation(int[] nums) {        if(nums == null || nums.length == 0) return;        int pivot = nums.length-1;        for(int i = nums.length-1; i >= 0; i--){            if(i >= 1 && nums[i] <= nums[i-1]) pivot = i-1;            else break;        }        if(pivot == 0){            reverse(nums, 0, nums.length-1);            return;        }        int idx = nums.length-1;        for(; idx > pivot-1; idx--){            if(nums[idx] > nums[pivot-1]) break;        }        int tmp = nums[pivot-1];        nums[pivot-1] = nums[idx];        nums[idx] = tmp;        reverse(nums, pivot, nums.length-1);    }    public void reverse(int[] nums, int start, int end){        if(start >= end) return;        for(int i = start; i <= (start + end)/2; i++){            swap(nums, i, start+end-i);        }    }    public void swap(int[] nums, int i, int j){        int tmp = nums[i];        nums[i] = nums[j];        nums[j] = tmp;    }}