目录 
题目
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
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.
解决方案
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| public class Solution {
public boolean isPalindrome(String s) {
int len = s.length();
if(s == null || s.length() == 0) return true;
StringBuilder builder1 = new StringBuilder();
StringBuilder builder2 = new StringBuilder();
s = s.toLowerCase();
boolean result = true;
for(int i = 0; i < len; i++){
if(isDigitOrLetter(s.charAt(i))){
builder1.append(s.charAt(i));
}
}
for(int j = len-1 ; j >= 0; j--){
if(isDigitOrLetter(s.charAt(j))){
builder2.append(s.charAt(j));
}
}
for(int k = 0; k < builder1.length(); k++){
if(builder1.charAt(k) != builder2.charAt(k)){
result = false;
break;
}
}
return result;
}
private static boolean isDigitOrLetter(char s){
if((s >= 'a' && s <= 'z') || (s >= 'A' && s <= 'Z') ||(s >= '0' && s <= '9')){
return true;
} else {
return false;
}
}
}
|
注意事项
- 先处理特殊情况，字符串为空或者长度为零。
- 定义一个函数isDigitOrLetter()来判断某个字符是否是字母大小写+数字。
- 使用两个StringBuilder，分别从首尾遍历字符串，存储为字母大小写+数字的字符。
- 比较两个StringBuilder中内容是否一样即可。
