# 31. Next Permutation

## 11 Jan 31. Next Permutation

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.

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

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]
```

Constraints:

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 100`

1　　2　　7　　4　　3　　1

1　　3　　1　　2　　4　　7

class Solution(object):
def nextPermutation(self, nums):
“””
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
“””
n = len(nums)
i = n – 1
while i > 0 and nums[i] <= nums[i – 1]:
i -= 1
self.reverse(nums, i, n – 1)
if i > 0:
for j in range(i, n):
if nums[j] > nums[i-1]:
self.swap(nums, i-1, j)
break

def reverse(self, nums, i, j):
“””
contains i and j.
“””
for k in range(i, (i + j) / 2 + 1):
self.swap(nums, k, i + j – k)

def swap(self, nums, i, j):
“””
contains i and j.
“””
nums[i], nums[j] = nums[j], nums[i]

C++代码如下：

class Solution {
public:
void nextPermutation(vector<int>& nums) {
const int N = nums.size();
int i = N – 2;
while (i >= 0 && nums[i] >= nums[i + 1]) –i;
int j = N – 1;
if (i >= 0) {
while (nums[j] <= nums[i]) –j;
swap(nums[i], nums[j]);
}
reverse(nums.begin() + i + 1, nums.end());
}
};
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class Solution {
public:
void nextPermutation(vector<int>& nums) {
next_permutation(nums.begin(), nums.end());
}
};

class Solution:
def nextPermutation(self, nums: List[int]) -> None:
“””
Do not return anything, modify nums in-place instead.
“””
n = len(nums)
i = n – 1
while i > 0 and nums[i]<=nums[i-1]:
i -= 1
self.revert(nums, i, n-1)
print(nums)
if i > 0:
for k in range(i, n):
if nums[k] > nums[i-1]:
self.swap(nums, k, i-1)
break
def revert(self, nums, i, j):
for k in range(i, (i+j)//2 +1 ):
self.swap(nums, k, i+j-k)

def swap(self, nums, i, j):
nums[i], nums[j] = nums[j], nums[i]