647. Palindromic Substrings

Medium
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.

看的答案，解法比较巧妙，遍历String的各个位置的字符串，从第i个字符串开始向外expand, 或者是从第i个字符串和第i+1个字符串向外expand, 如果一直满足s.charAt(l) == s.charAt(j),就一直 count++计 palindrome的数 。

class Solution {
    int count = 0;
    public int countSubstrings(String s) {
        int n = s.length();
        for (int i = 0; i < n; i++){
            expandPalindrome(s, i, i);
            expandPalindrome(s, i, i + 1);
        }
        return count;
    }
    
    private void expandPalindrome(String s, int l, int r){
        while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)){
            count++;
            l--;
            r++;
        }  
    }
}

本题还有一种方法，跟Longest Palindrome Substring的做法很像，也是用dp

class Solution {
    public int countSubstrings(String s) {
        int res = 0;
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--){
            for (int j = i; j < n; j++){
                if (s.charAt(i) == s.charAt(j) && (( j - i < 3) || (dp[i + 1][j - 1]))){
                    dp[i][j] = true;
                    res++;
                }
            }
        }
        return res;
    }
}