描述
Given a sorted array of integers, 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].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
分析
可以先二分查找，再从中心向左右扩张。
但这样其实在某些极端情况下性能不佳，比如全是相同元素数组。可以用两次二分查找，一次找上界，一次找下界。
代码
先二分，后扩张
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| class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
low = 0
high = len(nums) - 1
found = False
while low <= high:
mid = low + (high - low) / 2
if nums[mid] == target:
found = True
break
elif nums[mid] > target:
high = mid - 1
else:
low = mid + 1
if not found:
return [-1, -1]
i = mid - 1
while i >= 0 and nums[i] == target:
i -= 1
j = mid + 1
while j < len(nums) and nums[j] == target:
j += 1
return [i + 1, j - 1]
|
两次二分
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| class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
lower = self.lowerBound(nums, target)
upper = self.upperBound(nums, target)
if lower >= len(nums) or nums[lower] != target:
return [-1, -1]
else:
return [lower, upper]
def upperBound(self, nums, target):
low = 0
high = len(nums) - 1
while low <= high:
mid = low + (high - low) / 2
if nums[mid] <= target:
low = mid + 1
else:
high = mid - 1
return high
def lowerBound(self, nums, target):
low = 0
high = len(nums) - 1
while low <= high:
mid = low + (high - low) / 2
if nums[mid] >= target:
high = mid - 1
else:
low = mid + 1
return low
|