31. Next Permutation

Description

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.

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

Submitted Code

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        p = len(nums) - 2   
        while p >= 0 and nums[p] >= nums[p+1]:  # 오른쪽에서 처음으로 증가하는 지점 찾기
            p -= 1
        
        if p < 0:                               # 다시 순열 처음으로 리셋
            nums.reverse()
            return

        q = len(nums) - 1
        while nums[q] <= nums[p]:               # 피벗 뒤에서 피벗과 스왑할 자리 찾기
            q -= 1

        nums[p], nums[q] = nums[q], nums[p]     # 스왑
        nums[p+1:] = reversed(nums[p+1:])       # 피벗 뒤 반대로 재정렬(내림차순 -> 오름차순)

Runtime: 0 ms | Beats 100.00%
Memory: 19.36 MB | Beats 55.56%

사전 순 정렬에서 그 다음 순서를 찾는 문제로, 현재 순열보다 큰 순열 중에서 가장 작은 순열을 찾아야 한다. 오른쪽에서 처음으로 증가하는 지점을 찾아 그 값을 조금 더 큰 값으로 바꾸고, 뒤쪽을 최소(오름차순)로 만들면 된다.

nums = [1,2,3]

    
 1  (2) [3] →  1   3  [2]
[1]  3  (2) → [2]  3   1  → [2]  1   3  
 2  [1] (3) →  2  [3]  1 
[2] (3)  1  → [3]  2   1  → [3]  1   2 
 3  [1] (2) →  3  [2]  1 
 3   2   1  →  1   2   3

Other Solutions

1st

class Solution:
    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

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

두 위치를 변경 후 뒤쪽을 reverse()하는 대신 포인터 두 개로 직접 뒤집는 방법이다.