原题地址：https://oj.leetcode.com/problems/search-for-a-range/
题意：
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].
解题思路：又是二分查找的变形。因为题目要求的时间复杂度是O(log n)。在二分查找到元素时，需要向前和向后遍历来找到target元素的起点和终点。
代码：
class Solution: # @param A, a list of integers # @param target, an integer to be searched # @return a list of length 2, [index1, index2] def searchRange(self, A, target): left = 0; right = len(A) - 1 while left <= right: mid = (left + right) / 2 if A[mid] > target: right = mid - 1 elif A[mid] < target: left = mid + 1 else: list = [0, 0] if A[left] == target: list[0] = left if A[right] == target: list[1] = right for i in range(mid, right+1): if A[i] != target: list[1] = i - 1; break for i in range(mid, left-1, -1): if A[i] != target: list[0] = i + 1; break return list return [-1, -1]