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ARTS #148

Algorithm

本周选择的算法题是：Next Permutation。

规则

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 considered permutations of arr: [1,2,3], [1,3,2], [3,1,2], [2,3,1].

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

Solution

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        if n <= 1: return

        index = None
        for i in range(n-2, -1, -1):
            if nums[i] < nums[i+1]:
                index = i
                break

        def reverse(start: int, end: int):
            """
            反转指定的区间
            """
            while start < end:
                nums[start], nums[end] = nums[end], nums[start]
                start, end = start+1, end-1

        if index is None:
            """
            当数组完全是降序时，反转成升序
            """
            return reverse(0, n-1)
        
        smallest = index+1
        for i in range(smallest+1, n):
            if nums[i] <= nums[index]:
                break
            else:
                smallest = i

        nums[index], nums[smallest] = nums[smallest], nums[index]
        reverse(index+1, n-1)

Review

How we built Appflowy with Flutter and Rust

这篇文章介绍了 AppFlowy 如何用 Flutter + Rust 实现 Notion 的替代，目前在 GitHub 的 star 数接近 20k，其中的基础层采用 Rust 实现：

阅读下来非常顺滑，而且分层很清晰~

Tip

一个 macOS 下的开源代码编辑器： CodeEdit，纯 Swift 开发~

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