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

巴特西

leetcode 【 Search for a Range 】python 实现

最新文章

热门文章