LeetCode刷题之Next Permutation

Problem

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

My Solution

class Solution {
    public static int[] nums;

    public void nextPermutation(int[] nums) {
        int index = findIndex(nums);

        if (index == 0) {
            reverse(nums, index);
            return ;
        }

        exchangeHead(nums, index);
        reverse(nums, index);
    }

    public int findIndex(int[] nums) {
        for (int i = nums.length - 1; i > 0; --i) {
            if (nums[i] > nums[i - 1]) {
                return i;
            }
        }
        return 0;
    }

    public void exchangeHead(int[] nums, int index) {
        int head = nums[index - 1];
        for (int i = nums.length - 1; i >= 0; --i) {
            if (head < nums[i]) {
                nums[index - 1] = nums[i];
                nums[i] = head;
                break;
            }
        }
    }

    public void reverse(int[] nums, int index) {
        int p = index;
        for (int i = nums.length - 1; i > p; --i,++p) {
            int temp = nums[i];
            nums[i] = nums[p];
            nums[p] = temp;
        }
    }
}  

Great Solution

public void nextPermutation(int[] A) {
    if(A == null || A.length <= 1) return;
    int i = A.length - 2;
    while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
    if(i >= 0) {                           // If not entirely descending
        int j = A.length - 1;              // Start from the end
        while(A[j] <= A[i]) j--;           // Find rightmost first larger id j
        swap(A, i, j);                     // Switch i and j
    }
    reverse(A, i + 1, A.length - 1);       // Reverse the descending sequence
}

public void swap(int[] A, int i, int j) {
    int tmp = A[i];
    A[i] = A[j];
    A[j] = tmp;
}

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