[leetcode] 658. Find K Closest Elements @ python

最新推荐文章于 2021-08-12 21:54:08 发布

原创最新推荐文章于 2021-08-12 21:54:08 发布 · 343 阅读

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本文介绍了一种使用二分搜索法解决寻找已排序数组中k个最接近目标值x的元素的问题。通过确定起始索引，从该位置向两侧扩展以获取候选元素，并根据与目标值的距离进行调整，最终得到有序的结果。

原题

Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

Example 1:
Input: [1,2,3,4,5], k=4, x=3
Output: [1,2,3,4]
Example 2:
Input: [1,2,3,4,5], k=4, x=-1
Output: [1,2,3,4]
Note:
The value k is positive and will always be smaller than the length of the sorted array.
Length of the given array is positive and will not exceed 104
Absolute value of elements in the array and x will not exceed 104
UPDATE (2017/9/19):
The arr parameter had been changed to an array of integers (instead of a list of integers). Please reload the code definition to get the latest changes.

解法

二分搜索法. 使用bisect.bisect_left()确定index, 然后在index左右两边取k个数, 构成可能的升序序列, 然后对多余的数进行删除.

代码

class Solution:
    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
        i = bisect.bisect_left(arr, x)
        left = max(0,i-k)
        right = min(i+k-1, len(arr)-1)
        ans = arr[left:right+1]
        while len(ans) > k:
            # compare distance
            d_left = abs(ans[0] - x)
            d_right = abs(ans[-1] - x)
            if d_left > d_right:
                ans.pop(0)
            elif d_left <= d_right:
                ans.pop()
            
        return ans