This is a good question to ask during an interview. For the purpose of this problem, we define empty string as valid palindrome. SOLUTION 1: 左右指针往中间判断。注意函数: toLowerCase View Code SOLUTION 2: 引自http://blog.csdn.net/fightforyourdream/article/details/12860445的解答,会简单一点...
class Solution { public boolean isPalindrome(String s) { StringBuffer sgood = new StringBuffer(); int length = s.length(); for (int i = 0; i < length; i++) { char ch = s.charAt(i); if (Character.isLetterOrDigit(ch)) { sgood.append(Character.toLowerCase(ch)); } } String...
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: easy. AC代码: 1classSolution {2public:3boolisPalindrome(strings) {4inta=0,b=s.size()-1;5while(a<b){6if(!((s[a]>='a'&&s[a]<='z')||(s...
isValid(s[right])) { right--; continue; } if (s[left] === s[right]) { left++; right--; } else { break; } } return right <= left;};C++ Code:class Solution {public: bool isPalindrome(string s) { if (s.empty()) return true; const char*...
For the purpose of this problem, we define empty string as valid palindrome. 思路分析: 将字符串中的非数字字母字符跳过,大写全部转小写,然后从字符串两头向中间逼近,逐一进行比较。 C++参考示例: 代码语言:javascript 复制 classSolution{private:boolisAlphanumeric(char&ch){if(ch>='A'&&ch<='Z'){ch+...
class Solution: def validPalindrome(self, s: str) -> bool: # 定义左指针 l ,初始化为 0 l: int = 0 # 定义右指针 r ,初始化为 s.length - 1 r: int = len(s) - 1 # 当还有字符需要比较时,继续处理 while l < r: # 如果 s[l] 和 s[r] 不相等,则需要删除字符 if s[l] != ...
class Solution { public boolean isPalindrome(String s) { int st = 0; int ed = s.length() - 1; s = s.toLowerCase(); while (st < ed) { while (st < s.length() && !Character.isLetterOrDigit(s.charAt(st))) st++; while (ed >= 0 && !Character.isLetterOrDigit(s.charAt(ed...
For the purpose of this problem, we define empty string as valid palindrome. 程序如下: 1classSolution {2public:3boolisPalindrome(strings) {4if(!s.empty() && s.front() =='')5s.erase(0,1);6if(!s.empty() && s.back() =='')7s.pop_back();89boolret =true;10intsz =s.size()...
For the purpose of this problem, we define empty string as valid palindrome. Solution 代码语言:javascript 复制 classSolution{public:boolisPalindrome(string s){string res="";for(auto it:s){if((it>='a'&&it<='z')||(it>='0'&&it<='9'))res+=it;elseif(it>='A'&&it<='Z')res+=...
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