leetcode:5. Longest Palindromic Substring

一、问题描述

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.Example:Input: "cbbd"Output: "bb"

二、思路

• 思路一：暴力，判断每个字符串是不是回文串，若是回文串，若该串是目前最长的回文串，则更新最长长度与记录该回文串开始和结束的位置。具体代码看下面
• 思路二：一个长的回文串包含子回文串，暴力求解过程中重复去判断子回文串，导致了效率低下。一个好的方式，是从中心开始向外拓展回文串。具体代码看下面
• 思路三：使用Manacher算法，具体可以看以下两篇文章
  
  • Manacher’s ALGORITHM: O(n)时间求字符串的最长回文子串
  • 最长回文子串——Manacher 算法

三、代码

• 思路一代码：

public class Solution {    public String longestPalindrome(String s) {        int longest_length  = 0;        int begin = 0;        int end = 0;        int len = s.length();        for (int i = 0; i < len; i++){            for (int j = i; j < len; j++){                //更新最长回文串长度并记录开始和结束的位置                if (isPalindrome(s.substring(i, j + 1)) && (j - i + 1 ) > longest_length){                     begin = i;                    end = j + 1;                    longest_length = j - i + 1;                }            }        }        return s.substring(begin, end);    }    //判断是不是回文串    private static boolean isPalindrome(String s){        for (int i = 0; i < s.length() / 2; i++){            if (s.charAt(i) != s.charAt(s.length() - i - 1)){                return false;            }        }               return true;    }}

• 思路二代码：

public class Solution {    public String longestPalindrome(String s) {        int i, j;        int begin = 0;        int end = 0;        int max_len = 0;        int cur_len = 0;        int s_len = s.length();        for (i = 0; i < s_len; i++){                                // the middle point of palindrome            for (j = 0; (i - j >= 0) && (i + j < s_len); j++){      // if the length of palindrome is odd                if (s.charAt(i - j) != s.charAt(i + j)){                    break;                }                cur_len = j * 2 + 1;            }            if (cur_len > max_len){                max_len = cur_len;                begin = i - j + 1;                end = i + j;            }            for (j = 0; (i - j >= 0) && (i + j + 1 < s_len); j++){   // if the length of palindrome is even                    if (s.charAt(i - j) != s.charAt(i + j + 1)){                    break;                }                cur_len = j * 2 + 2;            }            if (cur_len > max_len){                max_len = cur_len;                begin = i - j + 1;                end = i + j + 1;            }        }        return s.substring(begin, end);    }}

• 思路三代码：

public class Solution {    public String longestPalindrome(String s) {        s =  formatString(s);        int[] radius = new int[s.length()];              // The left radius of palindrome in every position        int pos = 0;        int max_right = 0;        int max_len = 0;        int begin = 0;        int end = 0;        for (int i = 0; i < s.length(); i++){            if (i < max_right){                radius[i] = Math.min(radius[2 * pos - i], max_right - i);            } else {                radius[i] = 1;            }            while (i - radius[i] >= 0 && i + radius[i] < s.length() && (s.charAt(i - radius[i]) == s.charAt(i + radius[i]))){                 radius[i]++;            }            if (i + radius[i] - 1 > max_right){                max_right = i + radius[i] - 1;                pos = i;            }            if (radius[i] - 1 > max_len){                max_len = radius[i] - 1;                begin = i - (radius[i] - 1);                end = i + radius[i];            }        }        return s.substring(begin, end).replace("#", "");    }    // Format the original string like "abc" to "#a#b#c#" for example    private static String formatString(String s){        StringBuffer sb = new StringBuffer("#");        for (int i = 0; i < s.length(); i++){            sb.append(s.charAt(i) + "#");        }        return sb.toString();    }}