回文串、动态规划、中心扩散、Manacher 算法

https://leetcode-cn.com/problems/longest-palindromic-substring/

方法一：中心扩散法

class Solution {
    public String longestPalindrome(String s) {
         if(s.length()<2) return s;
        char[] c = s.toCharArray();
        int count;
        int point;
        int maxcount=0;
        int maxpoint=0;
        int size=0;
        int L=c.length;
        boolean flagissig=true;

        for (int i = 0; i < L; i++) {
            point = i;
            size = Math.min(i, L - i - 1);
            count = 0;

            for (int j = 0; j <= size; j++) {
                if (c[i - j] == c[i + j]) {
                    count++;
                    continue;
                } else {
                    break;
                }
            }

            if (count > maxcount) {
                maxcount = count;
                maxpoint = point;
            }
        }


//回文串是偶数的情况再来一次
        for (int i = 0; i < L; i++) {
            point = i;
            size = Math.min(i, L - i - 2);
            count = 0;
            for (int j = 0; j <= size; j++) {
                if (c[i - j] == c[i + j +1]) {
                    count++;
                    continue;
                } else {
                    break;
                }
            }

            if (count >= maxcount) {
                flagissig=false;
                maxcount = count;
                maxpoint = point;
            }
        }

        if(flagissig)
        return s.substring(maxpoint - maxcount + 1, maxpoint + maxcount);
        else
            return s.substring(maxpoint - maxcount + 1, maxpoint + maxcount+1);
    }
}

 

方法二：移动窗口法

public class jarvis {

    public static void main(String[] args) {
        System.out.println(longestPalindrome("sojrsalskjdef"));
    }

    public static String longestPalindrome(String s) {
        String result = null;
        char[] chars = s.toCharArray();
        for (int i = chars.length; i >= 1; i--) {
            //获取所有这个长度的char数组
            for (int a = 0; a <= chars.length - i; a++) {
                //判断是否是回文
                boolean flag = false;
                for (int j = 0; j < i / 2; j++) {
                    if (chars[a + j] == chars[a + i - 1 - j]) {

                    } else {
                        flag = true;  //不是回文
                    }
                }
                if (flag == false) {
                    result = s.substring(a, a + i);
                    return result;
                }
            }
        }
        return null;
    }
}