31. Next Permutation

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.

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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

Solution

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class Solution {
    public void nextPermutation(int[] nums) {
        int i = nums.length - 2;
        // [2,3,1,3,3]
        // i = 2
        while (i >= 0 && nums[i + 1] <= nums[i]) {
            i--;
        }
        
        if (i >= 0) {
            int j = nums.length - 1;
            while (nums[j] <= nums[i]) {
                j--;
            }
            swap(nums, i, j);
        }
        reverse(nums, i + 1);
    }
    
    void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
        
    void reverse(int[] nums, int start) {
        int i = start;
        int j = nums.length - 1;
        
        
        while (i < j) {
            // System.out.println("reverse " + i + " " + j);
            swap(nums, i, j);
            i++;
            j--;
        }
    }
}

Solution 2021-10-29

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class Solution {
    public void nextPermutation(int[] arr) {
        int n = arr.length;
        int index = -1;
        for (int i = n - 1; i > 0; i--) {
            if (arr[i - 1] < arr[i]) {
                index = i - 1;
                break;
            }
        }
        if (index != -1) {
            for (int j = n - 1; j > index; j--) {
                if (arr[j] > arr[index]) {
                    swap(arr, index, j);
                    break;
                }
            }
        }

        reverse(arr, index + 1, n);
    }
    
    void swap(int[] arr, int i, int j) {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }
    
    void reverse(int[] arr, int i, int j) {
        j--;
        while (i < j) {
            swap(arr, i , j);
            i++;
            j--;
        }
    }
}

Solution 2022-01-30

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class Solution {
    public void nextPermutation(int[] nums) {
        int n = nums.length;
        int index = -1;
        // 5 4 3 2 1
        for (int i = n - 1; i > 0; i--) {
           if (nums[i - 1] < nums[i]) {
               index = i - 1;
               break;
           } 
        }
        
        if (index != -1) {
            for (int j = n - 1; j > index; j--) {
                if (nums[j] > nums[index]) {
                    swap(nums, index, j);
                    break;
                }
            }
            
        }
        reverse(nums, index + 1);  
    }
    
    void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
    
    void reverse(int[] nums, int i) {
        int lo = i; 
        int hi = nums.length - 1;
        
        while (lo < hi) {
            swap(nums, lo, hi);
            lo++;
            hi--;
        }
    }
}