leetcode1930 长度为 3 的不同回文子序列

思路1：

枚举中间的字符。具体实现中使用了位运算的技巧。

实现1：

class Solution
{
public:
    int countPalindromicSubsequence(string s)
    {
        int n = s.length();
        vector<int> pre(n), suf(n);
        for (int i = 0; i < n; i++)
        {
            int tmp = (i == 0 ? 0: pre[i - 1]);
            pre[i] |= tmp; pre[i] |= 1 << s[i] - 'a';
        }
        for (int i = n - 1; i >= 0; i--)
        {
            int tmp = (i == n - 1 ? 0: suf[i + 1]);
            suf[i] |= tmp; suf[i] |= 1 << s[i] - 'a';
        }
        vector<int> res(26, 0);
        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            int P = (i == 0 ? 0: pre[i - 1]);
            int S = (i == n - 1 ? 0: suf[i + 1]);
            res[s[i] - 'a'] |= P & S;
        }
        for (int i = 0; i < 26; i++)
        {
            ans += __builtin_popcount(res[i]);
        }
        return ans;
    }
};

思路2：

枚举两侧的字符。

实现2：

class Solution
{
public:
    int countPalindromicSubsequence(string s)
    {
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < 26; i++)
        {
            int b = -1, e = -1;
            for (int j = 0; j < n; j++)
            {
                if (s[j] == char('a' + i))
                {
                    if (b == -1) b = j;
                    e = j;
                }
            }
            unordered_set<char> st;
            if (b != -1)
            {
                for (int i = b + 1; i < e; i++)
                {
                    st.insert(s[i]);
                }
            }
            ans += (int)st.size();
        }
        return ans;
    }
};