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Valid Palindrome leetcode java
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.
题解:
这道题的几个点,
一就是alphanumeric characters and ignoring cases,字母和数字,忽略大小写。
二就是考虑空字符串是否为回文,最好在面试时候问下面试官,这里是认为空字符串是回文。
因为忽略大小写,所以就统一为大写。
然后就判断当前检查字符是否符合范围,否则大小指针挪动。
如果发现有大小指针指向的值有不同的,就返回false,否则,继续检查。
最后返回true。
代码如下:
1 public static boolean isPalindrome(String s) {
2 if(s.length()==0)
3 return true;
4
5 s = s.toUpperCase();
6 int low1 = ‘A‘, high1 = ‘Z‘;
7 int low2 = ‘0‘, high2 = ‘9‘;
8 int low = 0, high = s.length()-1;
9
10 while(low < high){
11 if((s.charAt(low)<low1||s.charAt(low)>high1)
12 && (s.charAt(low)<low2||s.charAt(low)>high2)){
13 low++;
14 continue;
15 }
16
17 if((s.charAt(high)<low1||s.charAt(high)>high1)
18 && (s.charAt(high)<low2||s.charAt(high)>high2)){
19 high--;
20 continue;
21 }
22 if(s.charAt(low) == s.charAt(high)){
23 low++;
24 high--;
25 }else
26 return false;
27 }
28 return true;
29 }
2 if(s.length()==0)
3 return true;
4
5 s = s.toUpperCase();
6 int low1 = ‘A‘, high1 = ‘Z‘;
7 int low2 = ‘0‘, high2 = ‘9‘;
8 int low = 0, high = s.length()-1;
9
10 while(low < high){
11 if((s.charAt(low)<low1||s.charAt(low)>high1)
12 && (s.charAt(low)<low2||s.charAt(low)>high2)){
13 low++;
14 continue;
15 }
16
17 if((s.charAt(high)<low1||s.charAt(high)>high1)
18 && (s.charAt(high)<low2||s.charAt(high)>high2)){
19 high--;
20 continue;
21 }
22 if(s.charAt(low) == s.charAt(high)){
23 low++;
24 high--;
25 }else
26 return false;
27 }
28 return true;
29 }
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