Problem Link:

http://oj.leetcode.com/problems/valid-palindrome/

The following two conditions would simplify the problem:

  • only alphanumerci characters considered
  • ignoring cases

Given a string, we check if it is a valid palindrome by following rules:

  1. An empty string is a valid palindrome;
  2. Letter case should be ignored;
  3. Blank space should be ignored;
  4. The string is a valid palindrome only if s[x] == s[n-x] for x = 0,1,..., n/2

Therefore, we design the algorithm that scan the string from both the beginning and the end. Each iteration, we compare them to check if the string is a valid palindrome so far.

The algorithm will terminate if the two characters are not same; when all the characters in the string are compared, the algorithm will return True.

The following code is the python code accepted by oj.leetcode.com.

class Solution:
# @param s, a string
# @return a boolean
n0 = ord('0')
n9 = ord('9')
A = ord('A')
Z = ord('Z')
def isAlphanumeric(self, c):
x = ord(c)
return ( self.n0 <= x <= self.n9 or \
self.A <= x <= self.Z ) def isPalindrome(self, s):
"""
Scan the string from two ends of the string
"""
low = 0
high = len(s)-1
s = s.upper()
while low <= high:
if not self.isAlphanumeric(s[low]):
low += 1
elif not self.isAlphanumeric(s[high]):
high -= 1
elif s[low] == s[high]:
low += 1
high -= 1
else:
return False
return True

【LeetCode OJ】Valid Palindrome的更多相关文章

  1. 【LeetCode练习题】Valid Palindrome

    Valid Palindrome Given a string, determine if it is a palindrome, considering only alphanumeric char ...

  2. 【LeetCode OJ】Interleaving String

    Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 ...

  3. 【LeetCode OJ】Reverse Words in a String

    Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reve ...

  4. LeetCode OJ:Valid Palindrome(验证回文)

    Valid Palindrome Given a string, determine if it is a palindrome, considering only alphanumeric char ...

  5. 【LeetCode OJ】Palindrome Partitioning

    Problem Link: http://oj.leetcode.com/problems/palindrome-partitioning/ We solve this problem using D ...

  6. 【LeetCode OJ】Palindrome Partitioning II

    Problem Link: http://oj.leetcode.com/problems/palindrome-partitioning-ii/ We solve this problem by u ...

  7. 【LeetCode OJ】Gas Station

    Problem link: http://oj.leetcode.com/problems/gas-station/ We can solve this problem by following al ...

  8. 【LeetCode OJ】Word Break II

    Problem link: http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the w ...

  9. 【leetcode dp】132. Palindrome Partitioning II

    https://leetcode.com/problems/palindrome-partitioning-ii/description/ [题意] 给定一个字符串,求最少切割多少下,使得切割后的每个 ...

随机推荐

  1. Asp.net 解析json

    Asp.net Json数据解析的一种思路 http://www.cnblogs.com/scy251147/p/3317366.html http://tools.wx6.org/json2csha ...

  2. JSON生成c#类代码小工具(转)

    原文地址: http://www.cnblogs.com/tianqiq/archive/2015/03/02/4309791.html

  3. SqlServer中把结果集放到到临时表的方法

    一. SELECT INTO   1. 使用select into会自动生成临时表,不需要事先创建   select * into #temp from sysobjects   01. 把存储过程结 ...

  4. Objective-C:Foundation框架-常用类-NSNull

    集合中是不能存储nil值的,因为nil在集合中有特殊含义,但有时确实需要存储一个表示“什么都没有”的值,那么可以使用NSNull,它也是NSObject的一个子类. #import <Found ...

  5. 线性渐变--linear-gradient

    <!DOCTYPE html><html xmlns="http://www.w3.org/1999/xhtml"><head>    < ...

  6. Codeforces Round #308 (Div. 2)----C. Vanya and Scales

    C. Vanya and Scales time limit per test 1 second memory limit per test 256 megabytes input standard ...

  7. 319. Bulb Switcher——本质:迭代观察,然后找规律

    There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every ...

  8. Java中的String与常量池[转帖]

    string是java中的字符串.String类是不可变的,对String类的任何改变,都是返回一个新的String类对象.下面介绍java中的String与常量池. 1. 首先String不属于8种 ...

  9. NoSQL分类

    NoSQL数据库分类: NoSQL DEFINITION:Next Generation Databases mostly addressing some of the points: beingno ...

  10. 织梦dedecms分类信息模型上一页下一页失效办法

    修改文件/include/arc.archives.class 将一下代码 $next = (is_array($nextR) ? " where arc.id={$nextR['id']} ...