Palindromic Substrings

给一个字符串, 找到所有可能的回文子字符串的个数.

这题还是很巧妙的, 利用回文的性质, 就是奇数对称和偶数对称两种回文, 然后搜索, couting一下即可

class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        int res = 0;
        for(int i = 0; i < n; i++) {
            int odd = count(s, i, i);
            int even = count(s,i, i + 1);
            res += odd;
            res += even;
        }
        return res;
    }
    
    private int count(String s, int start, int end){
        int i = start;
        int j = end;
        int res = 0;
        while(i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)){
            res++;
            i--;
            j++;
        }
        return res;
    }
    //abbacca
}