[leetcode] 658. Find K Closest Elements @ python

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