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.

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

Follow up: Could you 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]

Analysis:

二分法找中间数，然后trace到第一个和最后一个。

improve：

两个二分法找lower和upper bound

Solution:

class Solution(object):
 def searchRange(self, nums, target):
 “””
 :type nums: List[int]
 :type target: int
 :rtype: List[int]
 “””
 if not nums: return [-1, -1]
 
 def lowerBound(nums, target):
   i, j = 0, len(nums)
   while i<j:
     mid = (i+j)/2
     if nums[mid] >= target:
       j = mid
     else:
       i = mid + 1
   return i
 
 def higherBound(nums, target):
   i, j = 0, len(nums)
   while i<j:
     mid = (i+j)/2
     if nums[mid] > target:
       j = mid
     else:
       i = mid + 1
   return i
 
 lower = lowerBound(nums, target)
 higher = higherBound(nums, target)-1
 if lower == len(nums) or nums[lower]!=target:
 return [-1, -1]
 
 return [lower, higher]