leetcode 1137. N-th Tribonacci Number(第N个泰波那契数)

The Tribonacci sequence Tn is defined as follows:

T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.

Given n, return the value of Tn.

Example 1:

Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
Example 2:

Input: n = 25
Output: 1389537

以前曾经见过Fibonacci数列,是前面两个数的和得到第3个数,
这里是Tribonacci数列,是前面3个数的和得到第4个数。
求第n个Tribonacci数。

思路:

DP
和求Fibonacci数一样,保存前面计算的结果到DP数组。

    public int tribonacci(int n) {
        if(n == 0) return 0;
        if(n <= 2) return 1;

        int[] dp = new int[n+1];
        dp[1] = 1;
        dp[2] = 1;

        for(int i = 3; i <= n; i++) {
            dp[i] = dp[i-3] + dp[i-2] + dp[i-1]; 
        }
        return dp[n];
    }

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