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34.find-first-and-last-position-of-element-in-sorted-array

Statement

Metadata

给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target,返回 [-1, -1]

进阶:

  • 你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗?

 

示例 1:

输入:nums = [5,7,7,8,8,10], target = 8
输出:[3,4]

示例 2:

输入:nums = [5,7,7,8,8,10], target = 6
输出:[-1,-1]

示例 3:

输入:nums = [], target = 0
输出:[-1,-1]

 

提示:

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • nums 是一个非递减数组
  • -109 <= target <= 109

Metadata

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

 

Constraints:

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • nums is a non-decreasing array.
  • -109 <= target <= 109

Solution

from bisect import bisect, bisect_left, bisect_right
from typing import List


class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        l = bisect_left(nums, target)
        r = bisect_right(nums, target)
        if l == len(nums) or nums[l] != target:
            return [-1, -1]
        return [l, r - 1]


if __name__ == "__main__":
    s = Solution()
    ans = s.searchRange([5, 7, 7, 8, 8, 10], 8)
    print(ans)

    ans = s.searchRange([1, 2, 3, 4, 5], 10)
    print(ans)

最后更新: October 11, 2023
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