leetcode 【 Search for a Range 】python 实现

题目：

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].

代码：oj测试通过 Runtime: 91 ms

 class Solution:
     # @param A, a list of integers
     # @param target, an integer to be searched
     # @return a list of length 2, [index1, index2]
     def searchAllTarget(self, A, index, target):
         # left index
         left_index = index
         curr_index = index
         while curr_index>=0 and A[curr_index]==target:
             left_index = curr_index
             curr_index = curr_index-1
         # right index
         right_index = index
         curr_index = index
         while curr_index<len(A) and A[curr_index]==target:
             right_index = curr_index
             curr_index = curr_index+1
         return [left_index,right_index]

     def searchRange(self, A, target):
         # none case
         if A is None:
             return None
         # short length cases
         if len(A)==1 :
             return[[-1,-1],[0,0]][A[0]==target]
         # binary search
         start = 0
         end = len(A)-1
         while start<=end :
             if start==end:
                 if A[start]==target :
                     return self.searchAllTarget(A, start, target)
                 else :
                     return [-1,-1]
             if start+1==end :
                 if A[start]==target :
                     return self.searchAllTarget(A, start, target)
                 elif A[end]==target :
                     return self.searchAllTarget(A, end, target)
                 else :
                     return [-1,-1]
             mid = (start+end)/2
             if A[mid]==target :
                 return self.searchAllTarget(A, mid, target)
             elif A[mid]>target :
                 end = mid-1
             else :
                 start = mid+1

思路：

这道题还是基于binary search，但是要求找到的是某个值的range。

分两步完成：

step1. 常规二分查找到target的某个index；如果没有找到则返回[-1,-1]

step2. 假设A中可能有多个位置为target，则从step1找到的index开始向左右search，直到把index左右两侧的target都找出来。

齐活儿