Leetcode: 31. Next Permutation

Question:

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 and use only constant 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

Note:

To find the next greater permutation, we need to search from the end of the array, and find the first descending element, switch it with the smallest element on it’s right. After this, we need to reserve the subarray on the right to make it arranged in increasing order.

Solution:

class Solution {
    public void nextPermutation(int[] nums) {
    	// find the first descending element
    	int i = nums.length - 2;
    	while (i >= 0 && nums[i] >= nums[i + 1]) {
    		i++;
    	}
    	// if the permutation is not already the largest
    	if (i >= 0) {
    		int j = nums.length - 1;
    		while (nums[j] <= nums[i]) {
    			j--;
    		}
    		int temp = nums[i];
    		nums[i] = nums[j];
    		nums[j] = temp;
    		// rearrange the right part of the array
    		reverse(nums, i+1);
    	} else {
    		reverse(nums, 0);
    	}
    }
    private void reverse(int[] arr, int start) {
    	for (int i = start, j = arr.length - 1; i < j; i++, j--) {
    		int temp = arr[i];
    		arr[i] = arr[j];
    		arr[j] = temp;
    	}
    }
}