【Leetcode】 1137. 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

Constraints:

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

---

Thought:

• 数组中套循环

class Solution {
public:
    int tribonacci(int n) {
        int* dp = new int[n + 1];
        dp[0] = 0;
        if(n >= 1) dp[1] = 1;
        if(n >= 2) dp[2] = 1;
        for(int i = 3; i <= n; i++)
        {
            dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
        }
        int res = dp[n];
        delete[] dp;
        return res;
    }
};

• 观察官方题解，发现方法一可以对我的代码的空间复杂度做出优化！

class Solution {
public:
    int tribonacci(int n) {
        if (n == 0) {
            return 0;
        }
        if (n <= 2) {
            return 1;
        }
        int p = 0, q = 0, r = 1, s = 1;
        for (int i = 3; i <= n; ++i) {
            p = q;
            q = r;
            r = s;
            s = p + q + r;
        }
        return s;
    }
};

显然，题解给出的滑动窗口更好！
方法二，事实上能用数学知识优化的代码效率更高效，
只可惜，鄙人学完了线性代数还是没有想到！

/*
 * @lc app=leetcode.cn id=1137 lang=cpp
 *
 * [1137] 第 N 个泰波那契数
 */

// @lc code=start
class Solution {
public:
    int tribonacci(int n) {
        if (n == 0) { // 如果n为0，返回0
            return 0;
        }
        if (n <= 2) { // 如果n小于等于2，返回1
            return 1;
        }
        vector<vector<long>> q = {{1, 1, 1}, {1, 0, 0}, {0, 1, 0}}; // 初始化矩阵q
        vector<vector<long>> res = pow(q, n); // 计算矩阵q的n次方
        return res[0][2]; // 返回矩阵q的(1,3)位置的值
    }

    vector<vector<long>> pow(vector<vector<long>>& a, long n) { // 计算矩阵a的n次方
        vector<vector<long>> ret = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; // 初始化单位矩阵
        while (n > 0) { // 当n大于0时
            if ((n & 1) == 1) { // 如果n为奇数
                ret = multiply(ret, a); // 计算ret和a的乘积
            }
            n >>= 1; // n右移一位
            a = multiply(a, a); // 计算a的平方
        }
        return ret; // 返回ret
    }

    vector<vector<long>> multiply(vector<vector<long>>& a, vector<vector<long>>& b) { // 计算矩阵a和矩阵b的乘积
        vector<vector<long>> c(3, vector<long>(3)); // 初始化结果矩阵c
        for (int i = 0; i < 3; i++) { // 遍历矩阵a的行
            for (int j = 0; j < 3; j++) { // 遍历矩阵b的列
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j]; // 计算c[i][j]的值
            }
        }
        return c; // 返回结果矩阵c
    }
};

// @lc code=end