【leetcode】Search for a Range

题目描述：

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：

明显的简单二分问题，首先用二分找到一个满足条件的点，然后向两边延展即可

coding=utf-8

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):l = len(A)left = 0 right = l-1index1 = -1index2 = -1pos = -1while left <= right:mid = (left + right) / 2if A[mid] == target:pos = midbreakelif A[mid] > target:right = mid - 1else:left = mid + 1#print posif pos == -1:return [-1,-1]index1 = index2 = poswhile A[index1] == target and index1 > 0 and A[index1-1] == target:index1 -= 1while A[index2] == target and index2 < l-1 and A[index2+1] ==target:index2 += 1return [index1,index2]s = Solution()
a = [5, 7, 7, 8, 8, 10]
print s.searchRange(a,8)
a = [1]
print s.searchRange(a,1)