1. 判断是不是回文串：
bool isPalindrome(const string& s) {
    int left = 0, right = s.length() - 1;
    while (left < right) {
        if (s[left] != s[right])
            return false;
        left ++;
        right --;
    }
    return true;
}
2. 动态
string removeShortestPalindrome(string s) {
    if (s.empty()) return s;

    int n = s.length();
    vector<vector<bool> > dp(n, vector<bool>(n, false));

    // 初始化长度为1和2的回文串
    for (int i = 0; i < n; i++) {
        dp[i][i] = true;
        if (i < n - 1 && s[i] == s[i + 1])
            dp[i][i + 1] = true;
    }

    // 动态规划填充dp表
    for (int len = 3; len <= n; len++) {
        for (int i = 0; i + len - 1 < n; i++) {
            int j = i + len - 1;
            if (s[i] == s[j] && dp[i + 1][j - 1])
                dp[i][j] = true;
        }
    }

    // 找到最短回文子串的起始位置和长度
    int start = 0, min_len = n;
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            if (dp[i][j] && (j - i + 1) % 2 == 0 && (j - i + 1) < min_len) {
                start = i;
                min_len = j - i + 1;
            }
        }
    }
    // 删除最短回文子串
    if (min_len == n) return ""; // 整个字符串都是回文串
    return s.substr(0, start) + s.substr(start + min_len);
}