# 0031. Next Permutation

https://leetcode.com/problems/next-permutation

## Description

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**](http://en.wikipedia.org/wiki/In-place_algorithm) 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`

## ac

```java
class Solution {
    public void nextPermutation(int[] nums) {
        int swapIndex = -1;
        int leftMin = Integer.MAX_VALUE;
        
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i] < nums[i+1]) {
                swapIndex = i;
                break;
            }
        }
        
        if (swapIndex < 0) {
            // Cannot find next permutation.
            Arrays.sort(nums);
            return;
        }
        
        for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] > nums[swapIndex]) {
                swap(nums, swapIndex, i);
                break;
            }
        }
        
        Arrays.sort(nums, swapIndex + 1, nums.length);
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

// 1) find swap point backwardly; 2) find target swap position backwardly; 3) sort the res;
// O(NlogN) time, O(1) space
```
