标签：
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].
很早之前就做的这道题，然而不明就理，这道题其实思路很简单,找到该元素出现的第一个位置和最后一个位置，两次二分。但是这两次在mid元素等于target时的操作不一样，寻找左元素时，则end = mid。右边界时start = end。代码如下：
class Solution(object): def searchRange(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ if not nums: return [-1,-1] l = 0 r = len(nums)-1 while l+1 < r: #find left mid = l + (r-l)/2 if nums[mid] == target: r = mid elif nums[mid] > target: r = mid else: l = mid if nums[l]==target: left = l elif nums[r]==target: left = r else: left = -1 l = 0 r = len(nums)-1 while l+1 < r: #find the right mid = l+(r-l)/2 if nums[mid] == target: l = mid elif nums[mid] > target: r = mid else: l = mid if nums[r]==target: right = r elif nums[l] == target: right = l else: right = -1 return [left,right]
标签：
原文地址：http://www.cnblogs.com/sherylwang/p/5493615.html
