https://leetcode.com/problems/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]
class Solution(object):
def findFirst(self, nums, target):
lo = 0
hi = len(nums) - 1
if hi < 0:
return -1
while lo <= hi:
mid = (lo + hi) / 2
if nums[mid] == target:
if mid == 0 or nums[mid] > nums[mid - 1]:
return mid
else:
hi = mid - 1
elif nums[mid] > target:
hi = mid - 1
else:
lo = mid + 1
return -1
def findLast(self, nums, target):
lo = 0
hi = len(nums) - 1
if hi < 0:
return -1
while lo <= hi:
mid = (lo + hi) / 2
if nums[mid] == target:
if mid == len(nums) - 1 or nums[mid] < nums[mid + 1]:
return mid
else:
lo = mid + 1
else:
hi = mid - 1
return -1
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
lo = self.findFirst(nums, target)
if lo == -1:
return [-1, -1]
hi = lo + self.findLast(nums[lo:], target)
return [lo, hi]