[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

Thought process:
Iterate through the array backwards. Find the first index i where nums[i] < nums[i + 1]. There are two cases:

1. i = -1: the array is originally in descending order. Reverse it.
2. i >= 0: 
1. Find the first element from the right that's larger than nums[i]. Swap it with nums[i].
2. We now need to fix the order of the elements to the right of i. From the iteration at the beginning, we know that all elements to the right of i are in descending order. Reverse those elements.

Solution:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

class Solution { public void nextPermutation(int[] nums) { int len = nums.length; int i = len - 2; while (i >= 0 && nums[i] >= nums[i + 1]) { i--; } if (i == -1) { reverse(nums, 0, len - 1); } else { int j = len - 1; while (nums[j] <= nums[i]) { j--; } swap(nums, i, j); reverse(nums, i + 1, len - 1); } } private void reverse(int[] nums, int left, int right) { while (left < right) { swap(nums, left, right); left++; right--; } } private void swap(int[] nums, int i, int j) { int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; } }

Time complexity: O(n).