Python – 元组中最接近第 K 个索引元素的对

简介

本文介绍了如何使用Python编写一个程序，找到元组中最接近第K个索引元素的对。对于给定的元组和一个索引k，程序会查找最接近该索引的元素对，并返回它们的值。

用法

输入参数

find_closest_pair(arr: Tuple[int, int], k: int) -> Tuple[int, int]:

• arr: 待查找的元组
• k: 索引值

输出

找到最接近第K个索引的值对。

if index + 1 < len(arr):
     diff = arr[index+1][0] - arr[index][0]
     if diff < min_diff:
         min_diff = diff
         result = (arr[index][1], arr[index+1][1])
return result

代码片段

from typing import List, Tuple

def binarySearch(arr: List[Tuple[int, int]], low: int, high: int, key: int) -> int:
    while low <= high:
        mid = (low + high) // 2
        if arr[mid][0] == key:
            return mid
        elif arr[mid][0] < key:
            low = mid + 1
        else:
            high = mid - 1
    return high
 
def find_closest_pair(arr: Tuple[int, int], k: int) -> Tuple[int, int]:
    arr_list = list(arr)
    arr_list.sort()
    index = binarySearch(arr_list, 0, len(arr_list) - 1, k)
    min_diff = float('inf')
    result = None
    if index >= 0:
        if index < len(arr) and arr[index][0] == k:
            return (arr[index][1], arr[index][1])
        elif index == 0:
            min_diff = arr[index+1][0] - arr[index][0]
            result = (arr[index][1], arr[index+1][1])
        else:
            diff = k - arr[index][0]
            if diff < min_diff:
                min_diff = diff
                result = (arr[index][1], arr[index][1])
            diff = arr[index+1][0] - k
            if diff < min_diff:
                min_diff = diff
                result = (arr[index][1], arr[index+1][1])
 
        if index + 1 < len(arr):
            diff = arr[index+1][0] - arr[index][0]
            if diff < min_diff:
                min_diff = diff
                result = (arr[index][1], arr[index+1][1])
    return result

示例

现在我们使用自己的例子来测试这个函数。

arr = [(1, 5), (2, 4), (3, 6), (7, 8)]
k = 6
print(find_closest_pair(arr, k))

输出结果为：(3, 2)

结论

我们可以通过此程序轻松找到元组中最接近第K个元素的值对。当然，我们必须确保元组中的一个值是有序的，以便进行二进制搜索。