125. Valid Palindrome
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.
判断字符串是否为回文字符串,只需考虑字符串中的数字和字母。
思路:字符串首尾各一个指针,进行比较;
指针移动规则:若相等则首指针向前移动一位,尾指针向后移动以为;若出现非法字符则首尾指针相应的移动一位。
代码如下:
1 class Solution { 2 public: 3 bool isPalindrome(string s) { 4 int n = s.size(); 5 int st = 0; 6 if(n == 0) 7 { 8 return true; 9 } 10 n--; 11 while(st < n) 12 { 13 if(isValid(s[st]) && isValid(s[n])) 14 { 15 if(s[st] <= 57 && s[st] >= 48) 16 { 17 if(s[n] > 57 || s[n] < 48) 18 { 19 20 return false; 21 } 22 else if(s[st] != s[n]) 23 { 24 return false; 25 } 26 else 27 { 28 st++; 29 n--; 30 continue; 31 } 32 } 33 if(s[st] == s[n] || s[st]-s[n] == 32 || s[n]-s[st] == 32) 34 { 35 st++; 36 n--; 37 continue; 38 } 39 else 40 { 41 return false; 42 } 43 } 44 else if(isValid(s[st])) 45 { 46 n--; 47 } 48 else if(isValid(s[n])) 49 { 50 st++; 51 } 52 else 53 { 54 n--; 55 st++; 56 } 57 } 58 return true; 59 } 60 bool isValid(char s) 61 { 62 if(s < 48 || s > 122) 63 { 64 return false; 65 } 66 else if(s > 57 && s < 65) 67 { 68 return false; 69 } 70 else if(s > 90 && s < 97) 71 { 72 return false; 73 } 74 return true; 75 } 76 };
时间: 2024-10-14 05:00:26