[LeetCode] Next Permutation

[LeetCode] Next Permutation
Next Permutation Total Accepted: 14635 Total Submissions: 57694 My Submissions
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

In three steps,

1. From the back to the front, find the first non-incremental number.

2. Find the first number that is larger than the previous number from the back and switch the position.

3. Reverse the sequence after the first number

public class Solution {    public void nextPermutation(int[] num) {        if (num == null || num.length < 2) {            return;        }                int len = num.length;                int i = len - 2;        for (; i >= 0; i--) {            if (num[i] < num[i + 1]) {                break;            }        }                if (i == -1) {            reverse(num, 0, len - 1);            return;        }                for (int j = len - 1; j > i; j--) {            if (num[j] > num[i]) {                swap(num, j, i);                break;            }        }                reverse(num, i + 1, len - 1);        return;    }        private void reverse(int[] num, int start, int end) {        while (start < end) {            swap(num, start, end);            start++;            end--;        }    }        private void swap(int[] num, int i, int j) {        int temp = num[i];        num[i]  = num[j];        num[j] = temp;    }}