Leet Code OJ 125. Valid Palindrome [Difficulty: Easy]
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题目:
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
翻译:
给定一个字符串,检测它是否是回文对称的,只考虑其中字母与数字的字符。
例如”A man, a plan, a canal: Panama”是回文对称的,”race a car”不是。
提示:你考虑过字符串可能是空的情况吗?这是一个面试中应该问出的好问题。为了处理这个问题,我们假定空串是回文对称的。
分析:
直接遍历比较是很困难的,我们需要先进行过滤(大写转小写)。过滤后只要将前后对应位置的字符直接比较就可以了。过程中要注意有的时候会涉及char和int的强转。
Java版代码:
public class Solution { public boolean isPalindrome(String s) { char[] charArr=s.toCharArray(); List<Integer> list=new ArrayList<>(); int fix='a'-'A'; for(char c:charArr){ if((c>='a'&&c<='z')||(c>='0'&&c<='9')){ list.add((int)c); }else if(c>='A'&&c<='Z'){ list.add(c+fix); } } int size=list.size(); for(int i=0;i<size/2;i++){ if(list.get(i)!=list.get(size-1-i)){ return false; } } return true; }}
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