Leetcode 647. Palindromic Substrings

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output:
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: "aaa"
Output:
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:

1. The input string length won't exceed 1000.

Similar Questions: Longest Palindromic Substring Longest Palindromic Subsequence Palindromic Substrings

Next challenges: Longest Palindromic Subsequence

方法一：O(n^2)的时间复杂度。

思路：回文字符串有奇数个字符时，当前i位置字母作为回文字符串的中心；有偶数个字符的时候当前i和i+1位置字母作为回文字符串的中心。

代码：

 public class Solution {
     public int countSubstrings(String s) {
         if(s == null) return 0;
         int count = 0;
         for(int i = 0; i < s.length(); i++) {
             count += countPalindromicSubstrings(s, i, i);
             count += countPalindromicSubstrings(s, i, i + 1);
         }
         return count;
     }

     public int countPalindromicSubstrings(String s, int begin, int end) {
         int count = 0;
         while(begin >= 0 && end < s.length() && s.charAt(begin) == s.charAt(end)) {
             count++;
             begin--;
             end++;
         }
         return count;
     }
 }

方法二：O(n)的时间复杂度。

思路：先对字符串进行改造（例如原字符串是"bab"，改造后是"#b#a#b#"），接着对改造后的字符串运行Manacher's Algorithm（“马拉车”算法），得到以s[i]为中心的回文串的半径RL[i]（不包括中心。例如"a"的半径就是0；"bab"以"a"为中心，半径就是1），显然，以s[i]为中心，RL[i]为半径的回文串中含有的字回文串数目是(RL[i] + 1) / 2个。最后只要将每个(RL[i] + 1) / 2加和就是结果。

关于Manacher's Algorithm的学习资料：

https://segmentfault.com/a/1190000003914228

http://www.cnblogs.com/grandyang/p/4475985.html

代码：

 public class Solution {
     public int countSubstrings(String s) {
         String rs = "#";
         //改造
         for(int i = 0; i < s.length(); i++) rs = rs + s.charAt(i) + "#";
         int[] RL = new int[rs.length()];//半径
         int pos = 0, maxRight = 0, count = 0;
         for(int i = 0; i < rs.length(); i++) {
             if(i < maxRight) {
                 RL[i] = Math.min(maxRight - i, RL[2 * pos - i]);
             }
             while(i - RL[i] - 1 >= 0 && i + RL[i] + 1< rs.length() && rs.charAt(i - RL[i] - 1) == rs.charAt(i + RL[i] + 1)) {
                 RL[i]++;
             }
             if(i + RL[i] > maxRight) {
                 pos = i;
                 maxRight = i + RL[i];
             }
             count += (RL[i] + 1) / 2;
         }
         return count;
     }
 }