[DP]1137. N-th Tribonacci Number

1137. N-th Tribonacci Number

Difficulty: 简单

The Tribonacci sequence T_n is defined as follows:

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

Given n, return the value of T_n.

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

Constraints:

• 0 <= n <= 37
• The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.

Solution 1

首先就是递归写法，结果意料中的运行超时
Language: c++

​class Solution {
public:
    int tribonacci(int n) {
        if(n == 0 || n == 1)
            return n;
        else if (n == 2)
            return 1;
        else{
            return tribonacci(n - 1) + tribonacci(n -2) + tribonacci(n-3);
        }
    }
};

由于限制个数最多到37，因此不出意料的超时。因此，记录下重复计算的项节约时间。

Solution 2

Language: c++

class Solution
{
public:
    int tribonacci(int n)
    {
        if (n == 0 || n == 1)
            return n;
        else if (n == 2)
            return 1;
        else
        {
            int a[50];
            a[0] = 0, a[1] = 1, a[2] = 1;
            //Tn+3 = Tn + Tn+1 + Tn+2
            for (int i = 3; i <= n; ++i)
            {
                a[i] = a[i - 1] + a[i - 2] + a[i - 3];
            }
            return a[n];
        }
    }
};

考虑到开辟了一次长度为50的数组，空间复杂度似乎可以进一步降低，因为每个值只与前三个数相关。因此整三个变量就行了。

Solution 3

Language: c++

class Solution
{
public:
    int tribonacci(int n)
    {
        if (n == 0 || n == 1)
            return n;
        else if (n == 2)
            return 1;
        else
        {
            int a = 0, b = 1, c = 1;
            //Tn+3 = Tn + Tn+1 + Tn+2
            for (int i = 3; i <= n; ++i)
            {
                c = a + b + c;
                b = c - a - b;
                a = c - a - b;
            }
            return c;
        }
    }
};