# Easiest JAVA Solution

• Using a simple example, we can derive the following steps:

``````public void nextPermutation(int[] A) {
if(A == null || A.length <= 1) return;
int i = A.length - 2;
while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
if(i >= 0) {                           // If not entirely descending
int j = A.length - 1;              // Start from the end
while(A[j] <= A[i]) j--;           // Find rightmost first larger id j
swap(A, i, j);                     // Switch i and j
}
reverse(A, i + 1, A.length - 1);       // Reverse the descending sequence
}

public void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}

public void reverse(int[] A, int i, int j) {
while(i < j) swap(A, i++, j--);
}``````

• Nice solution. I believe that there is no need to test for i < j in this line. It will always be satisfied. :

while(i < j && nums[j] <= nums[i]) j--;

• @rohan99 Nice catch! I've updated my answer.

• Hi @yavinci, great solution as always!

• @yanggao Thanks!

• there are two modifications I can think of:

1. adding a check for corner cases, since array may not have length >= 2, you may need:

`````` if (nums == null || nums.length <= 1) {
return;
}
``````
2. move this inside of the if block to make it more logical:

`` int j = nums.length - 1;``

• @yanggao, very nice catch. Updated. Thanks!

• This post is deleted!

• This is so coollll!

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