31. Next Permutation

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`

31. Next Permutation
``````struct Solution;

impl Solution {
fn next_permutation(nums: &mut Vec<i32>) {
let n = nums.len();
let mut i = n - 1;
while i > 0 && nums[i - 1] >= nums[i] {
i -= 1;
}
let mut j = i;
let mut k = n - 1;
while j < k {
nums.swap(j, k);
j += 1;
k -= 1;
}
if i > 0 {
k = i;
i -= 1;
while nums[k] <= nums[i] {
k += 1;
}
nums.swap(i, k)
}
}
}

#[test]
fn test() {
let mut nums = vec![1, 2, 3];
let res = vec![1, 3, 2];
Solution::next_permutation(&mut nums);
assert_eq!(nums, res);
let mut nums = vec![3, 2, 1];
let res = vec![1, 2, 3];
Solution::next_permutation(&mut nums);
assert_eq!(nums, res);
let mut nums = vec![1, 1, 5];
let res = vec![1, 5, 1];
Solution::next_permutation(&mut nums);
assert_eq!(nums, res);
}
``````