[Lintcode]61. Search for a Range/[Leetcode]34. Find First and Last Position of Element in Sorted Array
- 本题难度: Medium/Medium
Topic: Binary Search
Description
Given a sorted array of n integers, find the starting and ending position of a given target value.
If the target is not found in the array, return [-1, -1].
Example
Example 1:
Input:
[]
9
Output:
[-1,-1]
Example 2:
Input:
[5, 7, 7, 8, 8, 10]
8
Output:
[3, 4]
Challenge
O(log n) time.
我的代码
class Solution:
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
l = 0
r = len(nums) - 1
m = int((l + r) / 2)
r1 = -1
r2 = -1
while (l <= r):
if nums[m] == target and (m == 0 or nums[m - 1] < target):
r1 = m
r2 = m
l += 1
break
else:
if nums[m] >= target:
r = m - 1
else:
l = m + 1
m = int((l + r) / 2)
r = len(nums) - 1
while (l <= r):
if nums[m] == target and (m == r or nums[m + 1] > target):
r2 = m
break
else:
if nums[m] <= target:
l = m + 1
else:
r = m - 1
m = int((l + r) / 2)
return [r1,r2]别人代码
def searchRange(self, nums, target):
def search(n):
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) / 2
if nums[mid] >= n:
hi = mid
else:
lo = mid + 1
return lo
lo = search(target)
return [lo, search(target+1)-1] if target in nums[lo:lo+1] else [-1, -1]# 处理得很舒服def searchRange(self, nums, target):
lo = bisect.bisect_left(nums, target)
return [lo, bisect.bisect(nums, target)-1] if target in nums[lo:lo+1] else [-1, -1]思路
同类问题[Lintcode] 14. First Position of Target 但要注意，最后一个元素和第一个元素一个二分查找法是无法实现的。
- 时间复杂度: O(log(n))