LeetCode 1137.第 N 个泰波那契数(简单)

输入：n = 4
输出：4
解释：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

输入：n = 25
输出：1389537

class Solution {
    public int tribonacci(int n) {
        if (n == 0) {
            // f(0) = 0
            return 0;
        } else if (n == 1 || n == 2) {
            // f(1) = f(2) = 1
            return 1;
        }
        int p1 = 0, p2 = 1, cur = 1;
        for (int i = 3; i <= n; ++i) {
            int sum = p1 + p2 + cur;
            p1 = p2;
            p2 = cur;
            cur = sum;
        }
        return cur;
    }
}