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

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该博客围绕LeetCode 34题展开，要在升序整型数组中找目标值的起始和结束位置，不存在则返回[-1,-1]，要求时间复杂度不高于O(log n)。思路是用二分查找先找到首个目标值，再向两边扩展找边界，还给出Python代码，时间复杂度为O(log(n))。

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

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]

在一个升序整型数组中，找到目标值（多个）的起始位置和结束位置。如果不存在则返回[-1,-1]。要求时间复杂度不高于O(log n)。

思路

先用二分查找找到第一个目标值，然后向两边扩展找出所有目标值，并返回左右边界。

Python 代码实现

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        pos = -1
        left = 0
        right = len(nums)-1
        while left <= right:
            mid = (int)((left+right)/2)
            if nums[mid] == target:
                pos = mid
                break
            if nums[mid] > target :
                right = mid -1
            if nums[mid] < target :
                left = mid+1
                        
        if pos == -1:
            return [pos,pos]
        else:
            leftBound = pos
            rightBound = pos
            while leftBound >=0:
                if nums[leftBound] == target:
                    leftBound-=1
                else:
                    break
            while rightBound < len(nums):
                if nums[rightBound] == target:
                    rightBound+=1
                else:
                    break
            return [leftBound+1,rightBound-1]

时间复杂度：O(log(n))。

THE END.