认出名人：从 LeetCode 277 中学习如何找 ta

闲谈

2024-02-23 00:33:06

一、名人问题简介

在 LeetCode 的第 277 题中，我们需要解决一个有趣的问题：在一个派对中，有一个名人。名人与派对中的其他人不同，因为所有人都认识名人，但名人却不认识任何人。我们的任务是找出这位名人，并且只能询问两个人是否互相认识。

二、贪心算法

解决名人问题的贪心算法本质上是一个迭代过程，它从派对中的两个人开始，并根据他们的认识情况来逐步缩小名人候选人的范围。算法的核心步骤如下：

选择两个人，假设为 A 和 B。
如果 A 认识 B，那么 B 不可能是名人，因为名人不认识任何人。
如果 A 不认识 B，那么 A 不可能是名人，因为所有人都认识名人。
无论哪种情况，我们都可以排除掉其中一人。
重复以上步骤，直到只剩下一个人。这个人就是名人。

三、代码实现

以下是使用贪心算法解决名人问题的 Python 代码实现：

def find_celebrity(n):
    """
    Finds the celebrity in a party of n people.

    Args:
        n: The number of people in the party.

    Returns:
        The index of the celebrity, or -1 if no celebrity exists.
    """

    # Initialize two pointers, i and j, to the first two people in the party.
    i = 0
    j = 1

    # While both pointers are valid (i.e., within the range of 0 to n-1),
    # continue the loop.
    while i < n and j < n:
        # If i knows j, then j cannot be the celebrity.
        if knows(i, j):
            i = j + 1
        # If i does not know j, then i cannot be the celebrity.
        else:
            j = j + 1

    # After the loop, i points to a potential celebrity.
    # Check if i is indeed the celebrity by verifying that everyone knows i
    # and that i knows no one.
    celebrity = i
    for k in range(n):
        if k != celebrity and (not knows(k, celebrity) or knows(celebrity, k)):
            return -1

    return celebrity


def knows(a, b):
    """
    Returns True if person a knows person b, and False otherwise.

    Args:
        a: The index of person a.
        b: The index of person b.

    Returns:
        True if a knows b, and False otherwise.
    """

    # In a real-world scenario, this function would query a database or
    # perform some other operation to determine if a knows b.
    # For the purpose of this example, we assume that the knows() function
    # is implemented elsewhere and returns a boolean value.

    return True  # Replace this with the actual implementation of the knows() function.