leetcode第5题——**Longest Palindromic Substring

题目

Given a string  S , find the longest palindromic substring in  S . You may assume that the maximum length of  S  is 1000, and there exists one unique longest palindromic substring.

思路

题目是要求出字符串中的最长回文子串。定义计数变量cnt用于存储回文子串的长度，存储回文子串开始位置的变量start，光标cur，光标左右位置left和right。可以分为两种情况：1.连续相同字符如aabb这样的，遇到相同字符则cnt+1，然后光标移动开始下次循环；2.没有连续相同字符但是以某个字符对称如abcba，遇到对称字符则cnt+2，然后光标移动开始下次循环。最后substring(start,start+cnt)即为所求回文子串。

代码

Python

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        #cur是光标，maximum记录回文最大长度，cnt用于临时存储回文子串的长度,start是回文子串起始位置
        cur,maximum = 1,1
        cnt,start = 0,0
        if (len(s) == 2):
            if (s[0]==s[1]):
                return s
        while (cur < len(s)-1) & ((len(s)-cur-1)*2>maximum):
            left = cur - 1
            right = cur + 1
            cnt = 1
            if (s[left] == s[cur]) | (s[right] == s[cur]):
                while(left >= 0):
                    if s[left] == s[cur]:
                        cnt += 1
                        left -= 1
                    else:
                        break
                while(right  <= len(s)-1):
                    if s[right] == s[cur]:
                        cnt += 1
                        right += 1
                    else:
                        break
                cur = right
                while (left >= 0) & (right <= len(s)-1):
                    #处理类似asddddddsa这样的情况
                    if s[left] == s[right]:
                        cnt += 2
                        left -= 1
                        right += 1
                    else:
                        break
                if cnt > maximum:
                    maximum = cnt
                    start = left + 1
            else:
                #处理类似abcdcba这样以某字符对称的回文情况
                while (left >= 0) & (right <= len(s)-1):
                    #处理类似asddddddsa这样的情况
                    if s[left] == s[right]:
                        cnt += 2
                        left -= 1
                        right += 1
                    else:
                        break
                cur += 1
                if cnt > maximum:
                    maximum = cnt
                    start = left + 1
        return s[start:start+maximum]

Java

public class Solution {
public String longestPalindrome(String s) {
    char[] str=s.toCharArray();
    int max=1;//record max length of palindrome
    int cur=1;
    int left,right;
    int cnt;//record temp max length of palindrome
    int start=0;//record longest palindrome beginning location 
    if(str.length==2&&str[0]==str[1]) return s;
    while(cur<str.length-1&&(str.length-cur-1)*2>max){//处理长度大于三的字符串
        left=cur-1;
        right=cur+1;
        cnt=1;
        if(str[left]==str[cur]||str[right]==str[cur]){//判断是否有连续相同的字符
            while(left>=0){
                if(str[left]==str[cur]){
                    cnt++;
                    left--;
                }
                else break;
            }
            while(right<=str.length-1){
                if(str[right]==str[cur]){
                    cnt++;
                    right++;
                }
                else break;
            }
           cur=right;//reset cur location
           while(left>=0&&right<=str.length-1){// 类似'cbaaaaaabc'的情况
               if(str[left]==str[right]){
                   cnt=cnt+2;
                   left--;
                   right++;
               }
               else break;
           }
           if(cnt>max){
              max=cnt;
              start=left+1;// 最长回文子串的开始位置
           }
      }
      else{// 以光标为中心对称的回文字符串，如sdfgfdl
          while(left>=0&&right<=str.length-1){
              if(str[left]==str[right]){
                  cnt=cnt+2;
                  left--;
                  right++;
              }
              else break;
          }
          cur++;
          if(cnt>max){
              max=cnt;
              start=left+1;
          }
      }
    }
    return s.substring(start,start+max);
}
}