s0031_next_permutation

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 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

To solve the "Next Permutation" problem in Java with a Solution class, we can follow these steps:

1. Define a Solution class.
2. Define a method named nextPermutation that takes an integer array nums as input and modifies it to find the next permutation in lexicographic order.
3. Find the first index i from the right such that nums[i] > nums[i - 1]. If no such index exists, reverse the entire array, as it's already the last permutation.
4. Find the smallest index j from the right such that nums[j] > nums[i - 1].
5. Swap nums[i - 1] with nums[j].
6. Reverse the suffix of the array starting from index i.

Here's the implementation:

public class Solution {
    public void nextPermutation(int[] nums) {
        int n = nums.length;
        
        // Step 1: Find the first index i from the right such that nums[i] > nums[i - 1]
        int i = n - 1;
        while (i > 0 && nums[i] <= nums[i - 1]) {
            i--;
        }
        
        // Step 2: If no such index exists, reverse the entire array
        if (i == 0) {
            reverse(nums, 0, n - 1);
            return;
        }
        
        // Step 3: Find the smallest index j from the right such that nums[j] > nums[i - 1]
        int j = n - 1;
        while (nums[j] <= nums[i - 1]) {
            j--;
        }
        
        // Step 4: Swap nums[i - 1] with nums[j]
        swap(nums, i - 1, j);
        
        // Step 5: Reverse the suffix of the array starting from index i
        reverse(nums, i, n - 1);
    }
    
    // Helper method to reverse a portion of the array
    private void reverse(int[] nums, int start, int end) {
        while (start < end) {
            swap(nums, start, end);
            start++;
            end--;
        }
    }
    
    // Helper method to swap two elements in the array
    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

This implementation provides a solution to the "Next Permutation" problem in Java. It finds the next lexicographically greater permutation of the given array nums and modifies it in place.