# Next Permutation

Updated: Mar 25, 2021

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.

**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:**

```
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);
}
private void reverse(int[] nums, int start) {
int i = start, j = nums.length - 1;
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
// O(N)
}
```