LeetCode - 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: 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.

暴力解法， 注意要分情况讨论：

class Solution {
    public int countSubstrings(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        int count = 0;
        for(int i= 0; i<s.length(); i++){
            char c = s.charAt(i);
            //even case
            if(i < s.length() - 1){
                char d = s.charAt(i+1);
                if(c == d){
                    int j = 0;
                    while(i-j >= 0 && i+1+j < s.length() && s.charAt(i-j) == s.charAt(i+1+j)){
                        count++;
                        j++;
                    }
                }
            }
            //odd case
            int t = 0;
            while(i-t >= 0 && i+t < s.length() && s.charAt(i-t) == s.charAt(i+t)){
                count++;
                t++;
            }
            
        }
        return count;
    }
}

 二刷：

class Solution {
    public int countSubstrings(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        int count=0;
        for(int i = 0; i < s.length(); i++){
           count= count + help(s,i,i) + help(s, i, i+1);
           
        }
        return count;
    }
    
    public int help(String s, int i, int j){
        int count = 0;
        while(i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)){
            count++;
            i--;
            j++;
        }
        return count;
    }
}