[Lintcode]61. Search for a Range/[Leetcode]34. Find First and Last Position of Element in Sorted Array
- 本题难度: Medium/Medium
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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:
[5,7,8,10]
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,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,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))