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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0031_next_permutation;

// #Medium #Top_100_Liked_Questions #Array #Two_Pointers #Big_O_Time_O(n)_Space_O(1)
// #2023_08_09_Time_0_ms_(100.00%)_Space_42_MB_(90.28%)

/**
 * 31 - Next Permutation\.
 *
 * Medium
 *
 * Implement **next permutation** , which rearranges numbers into the lexicographically next greater permutation of numbers.
 *
 * If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).
 *
 * The replacement must be **[in place](http://en.wikipedia.org/wiki/In-place_algorithm)** and use only constant extra memory.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,2,3]
 *
 * **Output:** [1,3,2] 
 *
 * **Example 2:**
 *
 * **Input:** nums = [3,2,1]
 *
 * **Output:** [1,2,3] 
 *
 * **Example 3:**
 *
 * **Input:** nums = [1,1,5]
 *
 * **Output:** [1,5,1] 
 *
 * **Example 4:**
 *
 * **Input:** nums = [1]
 *
 * **Output:** [1] 
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 100`
 * *   `0 <= nums[i] <= 100`
**/
public class Solution {
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length <= 1) {
            return;
        }
        int i = nums.length - 2;
        while (i >= 0 && nums[i] >= nums[i + 1]) {
            i--;
        }
        if (i >= 0) {
            int j = nums.length - 1;
            while (nums[j] <= nums[i]) {
                j--;
            }
            swap(nums, i, j);
        }
        reverse(nums, i + 1, nums.length - 1);
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }

    private void reverse(int[] nums, int i, int j) {
        while (i < j) {
            swap(nums, i++, j--);
        }
    }
}