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].
题意：在一个有序数组中 查找某个target的范围，返回起始索引。 如果查找不到，返回[-1，-1]。要求算法复杂度logn。
直接二分查找到target的索引，然后往前后判断是否等于target。
class Solution(object): def searchRange(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ l,r=0,len(nums)-1 t=-1 a=[] if r==-1:return [-1,-1] while(l<=r): m=(l+r)/2 if(nums[m]==target): t=m break if(nums[m]>target):r=m-1 else:l=m+1 if t==-1:return [-1,-1] while(m>=0): if(nums[m]!=target): break m=m-1 a.append(m+1) while(t<=(len(nums)-1)): if(nums[t]!=target): break t=t+1 a.append(t-1) return a