Lintcode415-Valid Palindrome-Medium

最新推荐文章于 2022-05-06 00:12:08 发布

weixin_30784141

最新推荐文章于 2022-05-06 00:12:08 发布

阅读量100

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Jessiezyr/p/10643865.html

本文介绍了一种高效算法，用于判断字符串（仅考虑字母和数字）是否为回文。通过双指针法实现O(n)时间复杂度，并考虑了空字符串等特殊情况。

Given a string, determine if it is a palindrome, considering only alphanumeric(字母和数字) characters and ignoring cases.

Example

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama"

Example 2:

Input: "race a car"
Output: false
Explanation: "raceacar"

Notice

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.

思路：

字符串中只考虑字母和数字是否构成一个回文串，no extra space可以定义两个指针。

注意：

两个指针命名，front 和 end，不要用i, j
同类中：方法与方法之间直接用方法名来互相调用。Java方法的调用。
多个if并列和 else if 区别：If you have used multiple if statements then if the condition is true all will be executed. If you have used if and else if combination only one will be executed where first comes the true value
所以，如果把line 10 12 14 变成并列if，output会出错。因为如果连续出现两个或以上的非字母数字字符的话，就会输出false。比如 String s = a'+ba;
boolean方法命名以is开头。
line 9, 不能写成 while (front != end) , 举例 s = "aa", 指针front, end 交叉后会outOfIndexException.

代码：

 1 public class Solution {
 2     
 3     public boolean isPalindrome(String s) {
 4         if (s == null || s.length() == 0)
 5             return true;
 6             
 7         int front = 0;
 8         int end = s.length() - 1;
 9         while (front < end) {
10             if (!isValid(s.charAt(front)))
11                 front++;
12             else if (!isValid(s.charAt(end)))
13                 end--;
14             else {
15                 if (isValidCase(s.charAt(front), s.charAt(end))){
16                     front++;
17                     end--;
18                 } else {
19                     return false;
20                 }
21             }
22             
23         }
24         return true;
25     }
26     
27     public boolean isValid(char c) {
28         return Character.isLetter(c) || Character.isDigit(c);
29     }
30     public boolean isValidCase(char a, char b) {
31         if (a == b)
32             return true;
33         else if (Character.toUpperCase(a) == Character.toUpperCase(b))
34             return true;
35         else
36         return false;
37     }
38 }

代码优化：

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return true;
        }

        int front = 0;
        int end = s.length() - 1;
        while (front < end) {
            while (front < s.length() && !isvalid(s.charAt(front))) { // nead to check range of a/b
                front++;
            }

            if (front == s.length()) { // for empty string “.,,,”     
                return true; 
            }           

            while (end >= 0 && ! isvalid(s.charAt(end))) { // same here, need to check border of a,b
                end--;
            }

            if (Character.toLowerCase(s.charAt(front)) != Character.toLowerCase(s.charAt(end))) {
                return false;
            } else {
                front++;
                end--;
            }
        }

        return true; 
    }

    private boolean isvalid (char c) {
        return Character.isLetter(c) || Character.isDigit(c);
    }
}