31. 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

Solution:

From back to front, find the first index i that nums[i] < nums[i + 1].

From back to i, find the first index j that nums[j] > nums[i].

Swap nums[i] and nums[j].

Reverse nums[i + 1, end].

The time complexity is O(n) and the space complexity is O(1).



Code:

public class Solution {
    public void nextPermutation(int[] nums) {
        int i = nums.length - 2;
        while (i >= 0 && nums[i + 1] <= nums[i]) {
            i--;
        }
        if (i >= 0 ) {
            int j = nums.length - 1;
            while (j >= 0 && nums[j] <= nums[i]) {
                j--;
            }
            swap(nums, i, j);
        }
        reverse(nums, i + 1, nums.length - 1);
    }
    public void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
    public void reverse(int[] nums, int i, int j) {
        while (i < j) {
            swap(nums, i++, j--);
        }
    }
}