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

 1 class Solution:
 2     # @param A, a list of integers
 3     # @param target, an integer to be searched
 4     # @return a list of length 2, [index1, index2]
 5     def searchAllTarget(self, A, index, target):
 6         # left index
 7         left_index = index
 8         curr_index = index
 9         while curr_index>=0 and A[curr_index]==target:
10             left_index = curr_index
11             curr_index = curr_index-1
12         # right index
13         right_index = index
14         curr_index = index
15         while curr_index<len(A) and A[curr_index]==target:
16             right_index = curr_index
17             curr_index = curr_index+1
18         return [left_index,right_index]
19
20     def searchRange(self, A, target):
21         # none case
22         if A is None:
23             return None
24         # short length cases
25         if len(A)==1 :
26             return[[-1,-1],[0,0]][A[0]==target]
27         # binary search
28         start = 0
29         end = len(A)-1
30         while start<=end :
31             if start==end:
32                 if A[start]==target :
33                     return self.searchAllTarget(A, start, target)
34                 else :
35                     return [-1,-1]
36             if start+1==end :
37                 if A[start]==target :
38                     return self.searchAllTarget(A, start, target)
39                 elif A[end]==target :
40                     return self.searchAllTarget(A, end, target)
41                 else :
42                     return [-1,-1]
43             mid = (start+end)/2
44             if A[mid]==target :
45                 return self.searchAllTarget(A, mid, target)
46             elif A[mid]>target :
47                 end = mid-1
48             else :
49                 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