LeetCode05:Longest Palindromic Substring

原题目：

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

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

给定一个字符串，求出该字符串最长的回文子串。

回文串就是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。回文子串，顾名思义，即字符串中满足回文性质的子串。比如输入字符串 "google”，由于该字符串里最长的对称子字符串是 "goog”，因此输出4。

算法分析：

算法2：马拉车算法，时间复杂度：O(N),空间复杂度：O(N)

详见马拉车算法的博文

算法3：动态规划算法，时间复杂度：O(N^2),空间复杂度：O(N^2)

此题还可以用动态规划Dynamic Programming来解，我们维护一个二维数组dp，其中dp[i][j]表示字符串区间[i, j]是否为回文串，当i = j时，只有一个字符，肯定是回文串，如果i = j + 1，说明是相邻字符，此时需要判断s[i]是否等于s[j]，如果i和j不相邻，即i - j >= 2时，除了判断s[i]和s[j]相等之外，dp[j + 1][i - 1]若为真，就是回文串，通过以上分析，可以写出递推式如下：

           dp[i, j] = 1                                               if i == j

           = s[i] == s[j]                                if j = i + 1

           = s[i] == s[j] && dp[i + 1][j - 1]    if j > i + 1      

• dp[i][j] 表示子串s[i…j]是否是回文
  
• 初始化：dp[i][i] = true (0 <= i <= n-1);  dp[i][i-1] = true (1 <= i <= n-1); 其余的初始化为false
  
• dp[i][j] = (s[i] == s[j] && dp[i+1][j-1] == true)
  

遇到问题：

有点多啊，慢慢看程序吧

LeetCode提交源码：

算法2：

public String longestPalindrome(String s) {
        StringBuilder stringBuilder = new StringBuilder("$");
        for(char c : s.toCharArray()){
        	stringBuilder.append("#");
        	stringBuilder.append(c);
        }
        stringBuilder.append("#");
        String str = stringBuilder.toString();
        System.out.println("str=" + str);
        
        int center = 0;   			//中心位置
        int rightMax = 0;
        int index = 0;
        int maxLength = 0;
        
        int[] p = new int[str.length()];
        
        for(int i = 1;i < str.length();i++){
        	int j = 2*center -i;
        	
        	//p[i] = rightMax > i ? Math.min(p[symmetryId],rightMax - i) : 1;
        	if(rightMax > i){
        		if(p[j] < rightMax - i)
        			p[i] = p[j];
        		else
        			p[i] = rightMax - i;
        	}else{
        		p[i] = 1;
        		
        	}
        	
        	while (i + p[i] < str.length() && str.charAt(i + p[i]) == str.charAt(i - p[i])) {
                p[i]++;
            }
            if (i + p[i] > rightMax) {
                center = i;
                rightMax = i + p[i];
            }
            if (p[i] > maxLength) {
                maxLength = p[i];
                index = i;
            }
        }
        return s.substring((index - maxLength) / 2, (index + maxLength) / 2 - 1);
    }

 

LeetCode提交源码：

算法4：

public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        
        int length = s.length();    
        int max = 0;
        String result = "";
        for(int i = 1; i <= 2 * length - 1; i++){
            int count = 1;
            while(i - count >= 0 && i + count <= 2 * length  && get(s, i - count) == get(s, i + count)){
                count++;
            }
            count--; // there will be one extra count for the outbound #
            if(count > max) {
                result = s.substring((i - count) / 2, (i + count) / 2);
                max = count;
            }
        }
        
        return result;
    }
    
    private char get(String s, int i) {
        if(i % 2 == 0)
            return '#';
        else 
            return s.charAt(i / 2);
    }

 

完整运行程序：

/**************************************************************      
* Copyright (c) 2016      
* All rights reserved.                   
* 版 本 号：v1.0                   
* 题目描述：Longest Palindromic Substring
* 			Given a string s, find the longest palindromic substring in s. 
* 			You may assume that the maximum length of s is 1000.
* 			Input: "babad"
*			Output: "bab"
*			Note: "aba" is also a valid answer.
*			给定一个字符串，求出其中的最长回文子串
* 输入描述：请输入一个包含回文的字符串：
*			aaaba
* 程序输出： str=$#a#a#a#b#a#
*			最长的回文串为：aaa
* 问题分析：无
* 算法描述：算法2：马拉车算法,时间复杂度：O(N),空间复杂度：O(N)
* 		       算法3：动态规划算法, 时间复杂度：O(N^2),空间复杂度：O(N^2)
* 完成时间：2016-11-13
***************************************************************/ 
package org.GuoGuoFighting.LeetCode05;
import java.util.Scanner;
class SolutionMethod1{
	/*public String longestPalindrome(String s) {
        if(s == null){
        	return null;
        }
        char[] ch = s.toCharArray();
        int i = 0;
        int j = 1;
        char[] newch = new char[ch.length];
        
        while(i < ch.length){
        	for(;j < ch.length; j++){
        		if(ch[i] == ch[j]){
        			
        			break;
        		}
        	}
        	
        }
    }*/
}
/*
	算法2：马拉车算法 Manacher's ALGORITHM
	时间复杂度：O(N)
	空间复杂度：O(N)
*/
class SolutionMethod2{
	public String longestPalindrome(String s) {
        StringBuilder stringBuilder = new StringBuilder("$");
        for(char c : s.toCharArray()){
        	stringBuilder.append("#");
        	stringBuilder.append(c);
        }
        stringBuilder.append("#");
        String str = stringBuilder.toString();
        System.out.println("str=" + str);
        
        int center = 0;   			//中心位置
        int rightMax = 0;
        int index = 0;
        int maxLength = 0;
        
        int[] p = new int[str.length()];
        
        for(int i = 1;i < str.length();i++){
        	int j = 2*center -i;
        	
        	//p[i] = rightMax > i ? Math.min(p[symmetryId],rightMax - i) : 1;
        	if(rightMax > i){
        		if(p[j] < rightMax - i)
        			p[i] = p[j];
        		else
        			p[i] = rightMax - i;
        	}else{
        		p[i] = 1;
        		
        	}
        	
        	while (i + p[i] < str.length() && str.charAt(i + p[i]) == str.charAt(i - p[i])) {
                p[i]++;
            }
            if (i + p[i] > rightMax) {
                center = i;
                rightMax = i + p[i];
            }
            if (p[i] > maxLength) {
                maxLength = p[i];
                index = i;
            }
        }
        return s.substring((index - maxLength) / 2, (index + maxLength) / 2 - 1);
    }
}
/*
算法3：动态规划算法
时间复杂度：O(N^2)
空间复杂度：O(N^2)
*/
class SolutionMethod3{
	public String longestPalindrome(String s) {
		if(s.length() == 0){
			return "";
		}
		if(s.length() == 1){
			return s;
		}
		
		
        /*int length = s.length();
        boolean[][] dp = new boolean[length][length];
        int resleft = 0;
        int resright = 0;
        int i,j;
        for( i = 0; i < s.length(); i++){
        	for( j = 0; j < length; j++){
        		if(i >=j){
        			dp[i][j] = true;
        		}
        		else{
        			dp[i][j] = false;
        		}
        	}
        }
        int maxLen = 1;
        for(int k = 1; k < s.length(); k++){
        	for( i = 0; k +i < length; i++){
        		j = i + k;
        		if(s.charAt(i) != s.charAt(j)){
        			dp[i][j] = false;
        		}
        		else{
        			dp[i][j] = dp[i+1][j-1];
        			if(dp[i][j]){
        				if(k + 1 > maxLen){
        					maxLen = k + 1;
        					resleft = i;
        					resright = j;
        				}
        			}
        		}
        	}
        }
        return s.substring(resleft,resright+1);*/
		int maxLength = 0;
        String longest = null;
        int length = s.length();
        boolean[][] table = new boolean[length][length];
        // 单个字符都是回文
        for (int i = 0; i < length; i++) {
            table[i][i] = true;
            longest = s.substring(i, i + 1);
            maxLength = 1;
        }
        // 判断两个字符是否是回文
        for (int i = 0; i < length - 1; i++) {
            if (s.charAt(i) == s.charAt(i + 1)) {
                table[i][i + 1] = true;
                longest = s.substring(i, i + 2);
                maxLength = 2;
            }
        }
        // 求长度大于2的子串是否是回文串
        for (int len = 3; len <= length; len++) {
            for (int i = 0, j; (j = i + len - 1) <= length - 1; i++) {
                if (s.charAt(i) == s.charAt(j)) {
                    table[i][j] = table[i + 1][j - 1];
                    if (table[i][j] && maxLength < len) {
                        longest = s.substring(i, j + 1);
                        maxLength = len;
                    }
                } else {
                    table[i][j] = false;
                }
            }
        }
        return longest;
    }
}
class SolutionMethod4{
	public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        
        int length = s.length();    
        int max = 0;
        String result = "";
        for(int i = 1; i <= 2 * length - 1; i++){
            int count = 1;
            while(i - count >= 0 && i + count <= 2 * length  && get(s, i - count) == get(s, i + count)){
                count++;
            }
            count--; // there will be one extra count for the outbound #
            if(count > max) {
                result = s.substring((i - count) / 2, (i + count) / 2);
                max = count;
            }
        }
        
        return result;
    }
    
    private char get(String s, int i) {
        if(i % 2 == 0)
            return '#';
        else 
            return s.charAt(i / 2);
    }
}
public class LongestPalindromicSubstring {
	public static void main(String[] args){
		Scanner scanner = new Scanner(System.in);
		System.out.println("请输入一个包含回文的字符串：");
		String str = scanner.nextLine();
		scanner.close();
		SolutionMethod2 solution2 = new SolutionMethod2();
		
		System.out.println("算法2输出的最长回文串为：" + solution2.longestPalindrome(str));
		
		
		SolutionMethod3 solution3 = new SolutionMethod3();
		System.out.println("算法3输出的最长回文串为：" + solution3.longestPalindrome(str));
		
		SolutionMethod4 solution4 = new SolutionMethod4();
		System.out.println("算法4输出的最长回文串为：" + solution4.longestPalindrome(str));
		
		
	}
}

程序运行结果：