feat: add solutions to lc problem: No.1137 (#1618)

No.1137.N-th Tribonacci Number

Author: yanglbme

solution/1100-1199/1137.N-th Tribonacci Number/README.md

@@ -52,6 +52,32 @@ T_4 = 1 + 1 + 2 = 4
 
 时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为给定的整数。
 
+**方法二：矩阵快速幂加速递推**
+
+我们设 $Tib(n)$ 表示一个 $1 \times 3$ 的矩阵 $\begin{bmatrix} T_n & T_{n - 1} & T_{n - 2} \end{bmatrix}$，其中 $T_n$, $T_{n - 1}$ 和 $T_{n - 2}$ 分别表示第 $n$ 个、第 $n - 1$ 个和第 $n - 2$ 个泰波那契数。
+
+我们希望根据 $Tib(n-1) = \begin{bmatrix} T_{n - 1} & T_{n - 2} & T_{n - 3} \end{bmatrix}$ 推出 $Tib(n)$。也即是说，我们需要一个矩阵 $base$，使得 $Tib(n - 1) \times base = Tib(n)$，即：
+
+$$
+\begin{bmatrix}
+T_{n - 1} & T_{n - 2} & T_{n - 3}
+\end{bmatrix} \times base = \begin{bmatrix} T_n & T_{n - 1} & T_{n - 2} \end{bmatrix}
+$$
+
+由于 $T_n = T_{n - 1} + T_{n - 2} + T_{n - 3}$，所以矩阵 $base$ 为：
+
+$$
+\begin{bmatrix}
+ 1 & 1 & 0 \\
+ 1 & 0 & 1 \\
+ 1 & 0 & 0
+\end{bmatrix}
+$$
+
+我们定义初始矩阵 $res = \begin{bmatrix} 1 & 1  & 0 \end{bmatrix}$，那么 $T_n$ 等于 $res$ 乘以 $base^{n - 3}$ 的结果矩阵中所有元素之和。使用矩阵快速幂求解即可。
+
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -67,6 +93,35 @@ class Solution:
         return a
 ```
 
+```python
+class Solution:
+    def tribonacci(self, n: int) -> int:
+        def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
+            m, n = len(a), len(b[0])
+            c = [[0] * n for _ in range(m)]
+            for i in range(m):
+                for j in range(n):
+                    for k in range(len(a[0])):
+                        c[i][j] = c[i][j] + a[i][k] * b[k][j]
+            return c
+
+        def pow(a: List[List[int]], n: int) -> List[List[int]]:
+            res = [[1, 1, 0]]
+            while n:
+                if n & 1:
+                    res = mul(res, a)
+                n >>= 1
+                a = mul(a, a)
+            return res
+
+        if n == 0:
+            return 0
+        if n < 3:
+            return 1
+        a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
+        return sum(pow(a, n - 3)[0])
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -86,6 +141,51 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int tribonacci(int n) {
+        if (n == 0) {
+            return 0;
+        }
+        if (n < 3) {
+            return 1;
+        }
+        int[][] a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
+        int[][] res = pow(a, n - 3);
+        int ans = 0;
+        for (int x : res[0]) {
+            ans += x;
+        }
+        return ans;
+    }
+
+    private int[][] mul(int[][] a, int[][] b) {
+        int m = a.length, n = b[0].length;
+        int[][] c = new int[m][n];
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                for (int k = 0; k < b.length; ++k) {
+                    c[i][j] += a[i][k] * b[k][j];
+                }
+            }
+        }
+        return c;
+    }
+
+    private int[][] pow(int[][] a, int n) {
+        int[][] res = {{1, 1, 0}};
+        while (n > 0) {
+            if ((n & 1) == 1) {
+                res = mul(res, a);
+            }
+            a = mul(a, a);
+            n >>= 1;
+        }
+        return res;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -104,6 +204,50 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int tribonacci(int n) {
+        if (n == 0) {
+            return 0;
+        }
+        if (n < 3) {
+            return 1;
+        }
+        vector<vector<ll>> a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
+        vector<vector<ll>> res = pow(a, n - 3);
+        return accumulate(res[0].begin(), res[0].end(), 0);
+    }
+
+private:
+    using ll = long long;
+    vector<vector<ll>> mul(vector<vector<ll>>& a, vector<vector<ll>>& b) {
+        int m = a.size(), n = b[0].size();
+        vector<vector<ll>> c(m, vector<ll>(n));
+        for (int i = 0; i < m; ++i) {
+            for (int j = 0; j < n; ++j) {
+                for (int k = 0; k < b.size(); ++k) {
+                    c[i][j] += a[i][k] * b[k][j];
+                }
+            }
+        }
+        return c;
+    }
+
+    vector<vector<ll>> pow(vector<vector<ll>>& a, int n) {
+        vector<vector<ll>> res = {{1, 1, 0}};
+        while (n) {
+            if (n & 1) {
+                res = mul(res, a);
+            }
+            a = mul(a, a);
+            n >>= 1;
+        }
+        return res;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -116,6 +260,51 @@ func tribonacci(n int) int {
 }
 ```
 
+```go
+func tribonacci(n int) (ans int) {
+	if n == 0 {
+		return 0
+	}
+	if n < 3 {
+		return 1
+	}
+	a := [][]int{{1, 1, 0}, {1, 0, 1}, {1, 0, 0}}
+	res := pow(a, n-3)
+	for _, x := range res[0] {
+		ans += x
+	}
+	return
+}
+
+func mul(a, b [][]int) [][]int {
+	m, n := len(a), len(b[0])
+	c := make([][]int, m)
+	for i := range c {
+		c[i] = make([]int, n)
+	}
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			for k := 0; k < len(b); k++ {
+				c[i][j] += a[i][k] * b[k][j]
+			}
+		}
+	}
+	return c
+}
+
+func pow(a [][]int, n int) [][]int {
+	res := [][]int{{1, 1, 0}}
+	for n > 0 {
+		if n&1 == 1 {
+			res = mul(res, a)
+		}
+		a = mul(a, a)
+		n >>= 1
+	}
+	return res
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -137,6 +326,96 @@ var tribonacci = function (n) {
 };
 ```
 
+```js
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var tribonacci = function (n) {
+    if (n === 0) {
+        return 0;
+    }
+    if (n < 3) {
+        return 1;
+    }
+    const a = [
+        [1, 1, 0],
+        [1, 0, 1],
+        [1, 0, 0],
+    ];
+    return pow(a, n - 3)[0].reduce((a, b) => a + b);
+};
+
+function mul(a, b) {
+    const [m, n] = [a.length, b[0].length];
+    const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            for (let k = 0; k < b.length; ++k) {
+                c[i][j] += a[i][k] * b[k][j];
+            }
+        }
+    }
+    return c;
+}
+
+function pow(a, n) {
+    let res = [[1, 1, 0]];
+    while (n) {
+        if (n & 1) {
+            res = mul(res, a);
+        }
+        a = mul(a, a);
+        n >>= 1;
+    }
+    return res;
+}
+```
+
+### **TypeScript**
+
+```ts
+function tribonacci(n: number): number {
+    if (n === 0) {
+        return 0;
+    }
+    if (n < 3) {
+        return 1;
+    }
+    const a = [
+        [1, 1, 0],
+        [1, 0, 1],
+        [1, 0, 0],
+    ];
+    return pow(a, n - 3)[0].reduce((a, b) => a + b);
+}
+
+function mul(a: number[][], b: number[][]): number[][] {
+    const [m, n] = [a.length, b[0].length];
+    const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            for (let k = 0; k < b.length; ++k) {
+                c[i][j] += a[i][k] * b[k][j];
+            }
+        }
+    }
+    return c;
+}
+
+function pow(a: number[][], n: number): number[][] {
+    let res = [[1, 1, 0]];
+    while (n) {
+        if (n & 1) {
+            res = mul(res, a);
+        }
+        a = mul(a, a);
+        n >>= 1;
+    }
+    return res;
+}
+```
+
 ### **PHP**
 
 ```php