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

Code

public class Solution {
    /**
     * @param nums: an array of integers
     * @return: return nothing (void), do not return anything, modify nums in-place instead
     */
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0) return;

        // step1: search the first nums[k] < nums[k+1] backward
        int k = -1;
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i] < nums[i + 1]) {
                k = i;
                break;
            }
        }
        // if current rank is the largest, reverse it to smallest, return
        if (k == -1) {
            reverse(nums, 0, nums.length - 1);
            return;
        }

        // step2: search the first nums[k] < nums[l] backward
        int l = nums.length - 1;
        while (l > k && nums[l] <= nums[k]) l--;

        // step3: swap nums[k] with nums[l]
        int temp = nums[k];
        nums[k] = nums[l];
        nums[l] = temp;

        // step4: reverse between k+1 and nums.length-1;
        reverse(nums, k + 1, nums.length - 1);
    }

    private void reverse(int[] nums, int lb, int ub) {
        for (int i = lb, j = ub; i < j; i++, j--) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
}

Note

感觉这题纯数学，有点奇怪，不太明白

Reference

• Yu
• Yuanbin