荷兰国旗问题

荷兰国旗问题 | Dutch National Flag Problem

定义 | Definition

荷兰国旗问题是一种经典的排序问题，由 Edsger W. Dijkstra 提出。问题的目的是将包含三种不同颜色元素的数组重新排列，使得相同颜色的元素彼此相邻，颜色的顺序通常为红、白、蓝。

The Dutch National Flag problem is a classic sorting problem proposed by Edsger W. Dijkstra. The objective is to rearrange an array containing three different types of elements so that elements of the same type are adjacent, with the order typically being red, white, and blue.

解决方法 | Solution

该问题可以通过一种称为“三路划分”的算法来解决。该算法使用三个指针：low、mid 和 high，它们分别代表当前的红色、白色和蓝色元素的边界。

The problem can be solved using an algorithm called the “Three-way Partitioning”. This algorithm uses three pointers: low, mid, and high, representing the boundaries of the red, white, and blue elements, respectively.

算法步骤 | Algorithm Steps

1. 初始化: 将 low 和 mid 初始化为 0，将 high 初始化为数组的最后一个元素索引。 Initialization: Initialize low and mid to 0, and high to the index of the last element of the array.

2. 遍历数组: 当 mid 小于等于 high 时，执行以下操作： Traverse the array: While mid is less than or equal to high, do the following:

   • 如果当前元素是红色（通常为 0），则交换 low 和 mid 的元素，并将 low 和 mid 都增加 1。 If the current element is red (typically 0), swap the elements at low and mid, and increment both low and mid.
   • 如果当前元素是白色（通常为 1），只需将 mid 增加 1。 If the current element is white (typically 1), just increment mid.
   • 如果当前元素是蓝色（通常为 2），交换 mid 和 high 的元素，并将 high 减 1。 If the current element is blue (typically 2), swap the elements at mid and high, and decrement high.

def dutch_national_flag(arr):
    low, mid = 0, 0
    high = len(arr) - 1
    while mid <= high:
        if arr[mid] == 0:  # 红色 | Red
            arr[low], arr[mid] = arr[mid], arr[low]
            low += 1
            mid += 1
        elif arr[mid] == 1:  # 白色 | White
            mid += 1
        else:  # 蓝色 | Blue
            arr[mid], arr[high] = arr[high], arr[mid]
            high -= 1

复杂度分析 | Complexity Analysis

• 时间复杂度: 该算法只需遍历数组一次，因此时间复杂度为 $O(n)$，其中 $n$ 为数组的长度。 Time Complexity: The algorithm only needs to traverse the array once, so the time complexity is $O(n)$, where $n$ is the length of the array.
• 空间复杂度: 该算法只使用了常数个额外空间，因此空间复杂度为 $O(1)$。 Space Complexity: The algorithm uses only a constant amount of extra space, so the space complexity is $O(1)$.

常见错误 | Common Mistakes

1. 误用指针: 错误地移动 low 或 high 指针可能会导致无法正确地划分元素。 Misuse of Pointers: Incorrectly moving the low or high pointers can lead to improper partitioning of elements.
2. 处理顺序错误: 忘记按顺序处理三种颜色可能会导致错误的最终排列。 Wrong Order of Handling: Forgetting to handle the three colors in the correct order can result in an incorrect final arrangement.