[LeetCode] 1137. N-th Tribonacci Number

Description

e 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

Constraints:

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

Analyse

类似于fibonacci，由两项的递推变成了三项，也就是tribonacci

T0 = 0

T1 = 1

T2 = 1

Tn+3 = Tn + Tn+1 + Tn+2

构造一个递推公式，直接用递归来做，写起来很简单，可是超时了，原因是很多数据重复计算了

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

    return tribonacci(n-3) + tribonacci(n-2) + tribonacci(n-1);
}

改用动态规划的思想，AC了

int tribonacci(int n) {
    int list[38] = {0, 1, 1,};
    if (n < 3) return list[n];

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

    return list[n];
}

但还是比LeetCode上其他代码要慢，看了下更快的代码，唯一的差别就是把数组换成了vector

int tribonacci(int n) {
    vector list(38);
    int *list = (int*)malloc(sizeof(int) * 38); // 速度和数组一样，也比vector慢
    list[0] = 0;
    list[1] = 1;
    list[2] = 1;

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

    return list[n];
}

转载于:https://www.cnblogs.com/arcsinw/p/11425701.html