[Leetcode]34. Find First and Last Position of Element in Sorted Array

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本文介绍了一种算法，用于在升序整数数组中查找特定目标值的开始和结束位置，确保算法的时间复杂度为O(log n)。通过两个循环分别确定目标值的第一个和最后一个出现位置，提供了详细的代码实现和思路解析。

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34. Find First and Last Position of Element in Sorted Array

Description

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

Your algorithm’s runtime complexity must be in the order of O(log n).

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

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]

我的代码

class Solution:
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        l = 0
        r = len(nums) - 1
        m = int((l + r) / 2)
        r1 = -1
        r2 = -1
        while (l <= r):
            if nums[m] == target and (m == 0 or nums[m - 1] < target):
                    r1 = m
                    r2 = m
                    l += 1
                    break
            else:
                if nums[m] >= target:
                    r = m - 1
                else:
                    l = m + 1
                m = int((l + r) / 2)
        r = len(nums) - 1
        while (l <= r):
            if nums[m] == target and (m == r or nums[m + 1] > target):
                r2 = m
                break
            else:
                if nums[m] <= target:
                    l = m + 1
                else:
                    r = m - 1
                m = int((l + r) / 2)
        return [r1,r2]