Leetcode: Palindromic Substrings

647. Palindromic Substrings Medium 1656 85 Favorite Share 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: 3 Explanation: Three palindromic strings: "a", "b", "c". Example 2: Input: "aaa" Output: 6 Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa". Note: The input string length won't exceed 1000.

 

DP:

1 class Solution { 2 public int countSubstrings(String s) { 3 if (s == null || s.length() == 0) return 0; 4 int res = 0; 5 boolean[][] dp = new boolean[s.length()][s.length()]; 6 for (int i = s.length() - 1; i >= 0; i --) { 7 for (int j = i; j < s.length(); j ++) { 8 if (s.charAt(i) == s.charAt(j) && (j - i <= 2 || dp[i + 1][j - 1])) { 9 dp[i][j] = true; 10 res ++; 11 } 12 } 13 } 14 return res; 15 } 16 }

Extend palindrome: (better)

1 public class Solution { 2 int count = 0; 3 4 public int countSubstrings(String s) { 5 if (s == null || s.length() == 0) return 0; 6 7 for (int i = 0; i < s.length(); i++) { // i is the mid point 8 extendPalindrome(s, i, i); // odd length; 9 extendPalindrome(s, i, i + 1); // even length 10 } 11 12 return count; 13 } 14 15 private void extendPalindrome(String s, int left, int right) { 16 while (left >=0 && right < s.length() && s.charAt(left) == s.charAt(right)) { 17 count++; left--; right++; 18 } 19 } 20 }

 

转载于:https://www.cnblogs.com/EdwardLiu/p/11612467.html