（字符串操作）LeetCode#5. Longest Palindromic Substring

• 题目： 求字符串的最长回文子串
• 难度： Medium
• 思路： 遍历字符串，以每个元素为回文中心，向两边扩展找相同元素
• 代码：
  方法一：

public class Solution {    public String longestPalindrome(String s) {       //思路：定义两个变量maxLen和start来存储最长回文长度以及起始位置       int len = s.length();       if(s == null || len == 1 || len == 0){           return s;       }       int maxLen = 0;       int start = 0;       for(int i = 0; i < len;){           if(len-i <= maxLen/2){               break;           }           //left和right始终指向子串的起始和结束位置           int left = i;           int right = i;           while(right < len-1 && s.charAt(right+1) == s.charAt(right)){               right++;           }           i = right+1;//下次一个驼着的起始位置           while(left > 0 && right < len-1 && s.charAt(left-1) == s.charAt(right+1)){               left--;               right++;           }           if(right-left+1 > maxLen){               maxLen = right-left+1;               start = left;           }       }       return s.substring(start,start+maxLen);    }}

方法二：

public class Solution {    public String longestPalindrome(String s) {       //思路：定义两个变量maxLen和start来存储最长回文长度以及起始位置       int len = s.length();       if(s == null || len == 1 || len == 0){           return s;       }       int maxLen = 0;       String result = "";       for(int i = 0; i < len; i++){           int left = 0;           int right = 0;           //第一种情况：以第i个元素为中心，向两侧寻找相同元素           left = i-1;            right = i+1;           while(left >=0 && right <= len-1){               if(s.charAt(left) == s.charAt(right)){                   left--;                   right++;               }else{                   break;               }//当s.charAt(left) != s.charAt(right)时，一定要break，否则就会出现死循环           }           //或者上面的while循环写成            //while(left >= 0 && right <= len-1 && s.charAt(left) == s.charAt(right)){            //  left--;            //   right++;           //}           if(right-left-1 > maxLen){               maxLen = right-left-1;               result = s.substring(left+1,right);           }           //第二种情况：以第i和第i+1为中心           left = i;           right = i+1;           while(left >= 0 && right <= len-1){               if(s.charAt(left) == s.charAt(right)){                   left--;                   right++;               }else{                   break;               }           }           if(right-left-1 > maxLen){               maxLen = right-left-1;               result = s.substring(left+1,right);           }       }       return result;    }}