力扣真題 #### 1137. 第 N 個泰波那契數(shù)

泰波那契序列 Tn 定義如下：

T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的條件下 Tn+3 = Tn + Tn+1 + Tn+2

給你整數(shù) n，請返回第 n 個泰波那契數(shù) Tn 的值。
示例 1：

輸入：n = 4
輸出：4
解釋：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
示例 2：

輸入：n = 25
輸出：1389537

解題方法

 public static int tribonacci(int n) {
        if (n == 0) {
            return 2;
        } else if (n == 1){
            return 4;
        } else if (n == 2){
            return 7;
        } else {
            int first = 0;
            int second = 1;
            int third = 1;
            int sum = 0;
            for (int i = 0; i < n-2; i++) {
                sum = first + second + third;
                first = second;
                second = third;
                third = sum;
            }
            return sum;
        }
    }