剑指 Offer II 020. 回文子字符串的个数

class Solution:
    def countSubstrings(self, s: str) -> int:
        t = '^#' + '#'.join(s) + '#$'
        n = len(t)
        p = [0 for _ in range(n)]
        pos, maxRight = 0, 0
        ans = 0
        for i in range(1, n - 1):
            p[i] = min(maxRight - i, p[2 * pos - i]) if maxRight > i else 1
            while t[i - p[i]] == t[i + p[i]]:
                p[i] += 1
            if i + p[i] > maxRight:
                maxRight = i + p[i]
                pos = i
            ans += p[i] // 2
        return ans

class Solution {
    public int countSubstrings(String s) {
        StringBuilder sb = new StringBuilder("^#");
        for (char ch : s.toCharArray()) {
            sb.append(ch).append('#');
        }
        String t = sb.append('$').toString();
        int n = t.length();
        int[] p = new int[n];
        int pos = 0, maxRight = 0;
        int ans = 0;
        for (int i = 1; i < n - 1; i++) {
            p[i] = maxRight > i ? Math.min(maxRight - i, p[2 * pos - i]) : 1;
            while (t.charAt(i - p[i]) == t.charAt(i + p[i])) {
                p[i]++;
            }
            if (i + p[i] > maxRight) {
                maxRight = i + p[i];
                pos = i;
            }
            ans += p[i] / 2;
        }
        return ans;
    }
}