Tuesday, June 11, 2013

Search for a range@leetcode

刷题必备书籍Cracking the Coding Interview: 150 Programming Questions and Solutions 

简历:The Google Resume: How to Prepare for a Career and Land a Job at Apple, Microsoft, Google, or any Top Tech Company
算法学习书籍:Introduction to Algorithms
编程珠玑:Programming Pearls (2nd Edition)
C++ 学习:The C++ Programming Language, 4th Edition
经典操作系统书籍,龙书:Operating System Concepts
创业:The Start-up of You: Adapt to the Future, Invest in Yourself, and Transform Your Career
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].
» Solve this problem

5 comments:

  1. Replies
    1. We do binary search here. So it should be O(logn)

      Delete
    2. 你这个解法显然是错的,worst case O(n), not O(logn)

      Delete
  2. Worst case is not O(logn). Assume all 1s

    ReplyDelete
  3. We can do binary search twice. One is to find the most left index of the target and the other is the most right index of that. In this way, the worst case should be O(logn) but the code is not clean.

    ReplyDelete

Leetcode 316. Remove Duplicate Letters

 这道题表面问的是如何删除重复,实际在问如何从多个字符选取一个保留,从而让整个字符串按升序排列。那么策略就是对于高顺位的字符比如‘a',就要选靠前位置的保留,而低顺位字符如’z'则应该尽量选取靠后位置保留。 算法大概思路:每看到一个字符,我们要决定是否保留 1. ...