leetcode 算法整理

一 字符串中的最大回文串(第5题)

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

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

1. 我的解法(accepted): 中心扩展

思路: 回文即代表有中心，一次遍历中，对于每个位置上的数字求一下最大的回文串即可，初始处理剪枝，合并掉同样的元素，如xxxxaaaxxxx，先常量级别把a合并掉; 遍历时候再次剪枝，遍历过的有一个maxLen，如果即将遍历的元素最大可能回文长度都不可能超过maxLen，不再遍历。

 public class Test1218 {

     public static void main(String[] args) {
         String str = "ababababa";
         System.out.println(longestPalindrome(str));

     }

     public static String longestPalindrome(String s) {
         if (s == null) {
             return null;
         }
         char[] chars = s.toCharArray();
         int length = chars.length;
         int maxLen = 0;
         String maxStr = "";

         for (int i = 0; i < length; i++) {

             // cut branch
             int possibleLength = getMaxPossibleLength(i, length);
             if (possibleLength < maxLen) {
                 continue;
             }

             String maxStrTmp = getMaxStrByIndex(i, chars);
             if (maxLen < maxStrTmp.length()) {
                 maxLen = maxStrTmp.length();
                 maxStr = maxStrTmp;
             }
         }
         return maxStr;
     }

     private static int getMaxPossibleLength(int index, int length) {
         int head = 0;
         int tail = length - 1;
         if (index == head || index == tail) {
             return 1;
         }
         int result1 = index - head;
         int result2 = tail - index;

         int min = result1 <= result2 ? result1 : result2;
         return min * 2 + 1;
     }

     private static String getMaxStrByIndex(int index, char[] chars) {
         StringBuilder sb = new StringBuilder(String.valueOf(chars[index]));
         int length = chars.length;
         int head = index - 1;
         int tail = index + 1;

         // middle deal
         while (true) {
             if (head >= 0 && chars[index] == chars[head]) {
                 sb.insert(0, String.valueOf(chars[head--]));
             } else if (tail <= length - 1 && chars[index] == chars[tail]) {
                 sb.append(String.valueOf(chars[tail++]));
             } else {
                 break;
             }
         }

         // besides deal
         while (true) {
             if (head < 0 || tail > length - 1) {
                 break;
             }
             if (head >= 0 && tail <= length - 1 && chars[head] == chars[tail]) {
                 sb.insert(0, String.valueOf(chars[head--]));
                 sb.append(String.valueOf(chars[tail++]));
                 continue;
             }
             break;

         }
         return sb.toString();
     }
 }

2. dp解法

思路: 设 p[i][j] 代表下标从i至j的子字符串是否是回文串，取值为boolean

转移方程 p[i][j] = p[i+1][j-1] && chars[i] == chars[j]

初始化的状态为 p[i][i] = true    p[i][i+1] = chars[i] == chars[i+1]

看下面手绘图理解一下，打勾的对角线p[i][i] 恒为true, 只有这一列是不够状态转移的，因为按照转移方程，必须要图示的↗️方向递推，那么要需要有圆圈的一列初始化，这列对应的是p[i][i+1]。 剩下的递推即可，比如图中的箭头栗子