5. Longest Palindromic Substring [最长回文子串]

描述

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000

给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000

例子

思路

• 方法1 龙卷风法，找龙眼

两种龙眼：一种bab，一种baab，即(i,i)和(i,i+1)

• 方法2 动态规划
  答案
• java

//方法1
class Solution {
    private int max_i=0,max_j=0;
    public void check(String s, int i, int j) {
        
        while(i>=0 && j<=s.length()-1){
            if(s.charAt(i)==s.charAt(j)) {
                i--;
                j++;
            }else break;
        }
        i=i+1;
        j=j-1;
        if(j-i>max_j-max_i) {
            max_i=i;
            max_j=j;
        }
        
        
    }
    
    public String longestPalindrome(String s) {
        if(s.length()==0) return "";
        for(int i=0; i<s.length(); i++) {
            check(s,i,i);
            check(s,i,i+1);
        }
        return s.substring(max_i, max_j+1);
    }
}
//方法2
class Solution {
    public String longestPalindrome(String s) {
        if(s.length()==0) return "";
        boolean[][] arr = new boolean[s.length()][s.length()];

        int max_i=0, max_j=0;
        //r为字符的长度
        for(int r=1; r<=s.length(); r++) {
            for(int i=0; i+r-1<s.length(); i++) {
                int j=i+r-1;
                //长度为1
                if(i==j) arr[i][j]=true;
                //长度为2
                else if(i+1==j && s.charAt(i)==s.charAt(j)) arr[i][j]=true;
                else if(i+1<=j-1 && s.charAt(i)==s.charAt(j)) arr[i][j]=arr[i+1][j-1];
                
                if(arr[i][j]==true && j-i>max_j-max_i){
                    max_i=i;
                    max_j=j;
                }
            }
        }
        return s.substring(max_i, max_j+1);
}

• python

*方法1*
class Solution:
    def check(self, s:string, a:int,b:int)->str:
        while a>=0 and b<len(s) and s[a]==s[b]:
            a-=1
            b+=1
        a+=1
        b-=1
        return s[a:b+1]

    def longestPalindrome(self, s: str) -> str:
        res = ""
        for i in range(len(s)):
            
            s2 = self.check(s,i,i)
            s3 = self.check(s,i,i+1)
            if len(res)<len(s2):
                res = s2
            if len(res)<len(s3):
                res = s3
        return res
*方法2*
    def longestPalindrome(self, s: str) -> str:
        res = ""
        dp=[[0]*len(s) for i in range(len(s))]
        for r in range(len(s)):#r=j-i 0~len-1 r为0时是一个元素，为1时为两个元素在一起
            for i in range(len(s)-r): #i+r<=len-1(最大下标)
                j=i+r
                if i==j:
                    dp[i][j]=1
                if i+1==j:
                    dp[i][j]=1 if s[i]==s[j] else 0
                   
                if i+2<=j:
                    dp[i][j]=dp[i+1][j-1] and s[i]==s[j]
                if dp[i][j] and j-i+1>len(res):
                    res=s[i:j+1]
       

• c++

*方法1*
class Solution {
private:
    string check(string s, int a, int b){
        while (a>=0 && b<=s.size() && s[a]==s[b])
        {
            a--;
            b++;
        }
        a++;
        b--;
        return s.substr(a,b-a+1);
    }
public:
    
    string longestPalindrome(string s) {
        string res("");
        for (int i=0; i<s.size(); i++)
        {
            string s2 = check(s,i,i);
            string s3 = check(s,i,i+1);
            if (res.size()<s2.size())
                res=s2;
            if (res.size()<s3.size())
                res=s3;
            
        }
        return res;
    }
    
};
*方法2*
class Solution {
public:
    string longestPalindrome(string s) {
        int L = s.size();
        string res("");
        vector<vector<int>> dp(L,vector<int>(L));//全是0的二维数组
        for(int r=0; r<L; r++)
            for (int i=0; i<L-r; i++)
            {
                int j=i+r;
                if (i==j)
                    dp[i][j]=1;
                if (i+1==j)
                    dp[i][j]=s[i]==s[j]?1:0;
                if (i+2<=j)
                    dp[i][j]=(dp[i+1][j-1] && s[i]==s[j])?1:0;
                if (dp[i][j] && j-i+1>res.size())
                    res = s.substr(i,j-i+1);
                
            }
        
        return res;
    }
};