# Next Permutation LeetCode Solution

Here, We see Next Permutation LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given an array of integers nums, find the next permutation of nums.

```A permutation of an array of integers is an arrangement of its members into a sequence or linear order.

For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].

The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).

For example, the next permutation of arr = [1,2,3] is [1,3,2].
Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.```
```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]
```

## Next Permutation Leetcode Solution C++

``````class Solution {
public:
void nextPermutation(vector<int>& nums) {
int n = nums.size(), k, l;
for (k = n - 2; k >= 0; k--) {
if (nums[k] < nums[k + 1]) {
break;
}
}
if (k < 0) {
reverse(nums.begin(), nums.end());
} else {
for (l = n - 1; l > k; l--) {
if (nums[l] > nums[k]) {
break;
}
}
swap(nums[k], nums[l]);
reverse(nums.begin() + k + 1, nums.end());
}
}
};
```Code language: C++ (cpp)```

## Next Permutation Leetcode Solution Java

``````class Solution {
public void nextPermutation(int[] nums) {
int dest = nums.length - 2;
for (; 0 <= dest && nums[dest] >= nums[dest + 1]; --dest)
;
if (0 <= dest) {
int target = nums.length - 1;
for (; nums[dest] >= nums[target]; --target)
;
int tmp = nums[dest];
nums[dest] = nums[target];
nums[target] = tmp;
}
for (int end = nums.length - 1; dest + 1 < end; ) {
int tmp = nums[++dest];
nums[dest] = nums[end];
nums[end--] = tmp;
}
}
}
```Code language: Java (java)```

## Next Permutation Leetcode Solution JavaScript

``````var nextPermutation = function(nums) {
const swap = (a, b) => [nums[a],nums[b]] = [nums[b],nums[a]]
let len = nums.length - 1, i, j
for (i = len - 1; nums[i] >= nums[i+1];) i--
for (let k = i+1; len > k; k++, len--) swap(k,len)
if (~i) {
for (j = i + 1; nums[i] >= nums[j];) j++
swap(i,j)
}
};
```Code language: JavaScript (javascript)```

## Next Permutation Leetcode Solution Python

``````class Solution(object):
def nextPermutation(self, nums):
i = j = len(nums)-1
while i > 0 and nums[i-1] >= nums[i]:
i -= 1
if i == 0:   # nums are in descending order
nums.reverse()
return
k = i - 1    # find the last "ascending" position
while nums[j] <= nums[k]:
j -= 1
nums[k], nums[j] = nums[j], nums[k]
l, r = k+1, len(nums)-1  # reverse the second part
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l +=1 ; r -= 1
```Code language: Python (python)```
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