LeetCode_34. Find First and Last Position of Element in Sorted Array 在有序数组中查找某数字的起始和终止下标

题目：

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]

代码：

class Solution(object):
    
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        
        n = len(nums)
        index1 = -1       #起始下标
        index2 = -1       #终止下标

        i = 0

        while i < n and nums[i] <= target :
            if nums[i] == target and index1 == -1:     #标记第一个下标
                index1 = i

            if nums[i] == target and index1 != -1 and i < n-1 and nums[i+1] != target :  #查找第二个下标，判断不是数组末尾的情况
                index2 = i

            if nums[i] == target and index1 != -1 and i == n-1 :    #查找第二个下标，判断是数组末尾的情况
                index2 = i

            i += 1

        return [index1, index2]

 

顺序查找的算法，比较好理解，但没有满足题目要求的时间复杂度。没有使用二叉树的查找算法。