题目描述
大家都知道斐波那契数列,现在要求输入一个整数n,请你输出斐波那契数列的第n项(从0开始,第0项为0)。 n<=39
解题思路
递推公式f(n)=f(n)=
当n=0=0,当n=0 当
n=1=1,当n=1
其他=f(n−1)+f(n−2)看到这大家很容易想起递归,课堂上老师讲递归的时候的经典例子。但是当n很大的时候,就会出现堆栈溢出。堆栈溢出的主要原因是,递归重复的计算太多,很多计算是可以避免的,用循环计算结果,显根据前两项算出第三项,以后每次都是这样计算。
代码实现
递归实现
public static int Fibonacci(int n) { if (n <= 1) return n; return Fibonacci(n - 1) + Fibonacci(n - 2); }
循环实现
public static int Fibonacci2(int n) { if (n <= 1) return n; int first = 0; int second = 1; int result = 0; for (int i = 2; i <= n; i++) { result = first + second; first = second; second = result; } return result; }
斐波那契数列求和
public static int FibonacciSum(int n) { if (n <= 1) return n; int first = 0; int second = 1; int temp = 0; int result = first + second; for (int i = 2; i <= n; i++) { temp = first + second; first = second; second = temp; result = result + temp; } return result; }
斐波那契数列求和,利用公式计算
public static int FibonacciSum2(int n) { if (n <= 1) return n; int first = 0; int second = 1; int temp = 0; for (int i = 2; i <= n; i++) { temp = first + second; first = second; second = temp; } int result = 2 * second + first - 1; //Sn = 2an + an - 1 - 1 return result; }
测试
[Fact] public void Test0() { Assert.Equal(0, Coding007.Fibonacci(0)); Assert.Equal(0, Coding007.Fibonacci2(0)); Assert.Equal(0, Coding007.FibonacciSum(0)); Assert.Equal(0, Coding007.FibonacciSum2(0)); } [Fact] public void Test1() { Assert.Equal(1, Coding007.Fibonacci(1)); Assert.Equal(1, Coding007.Fibonacci2(1)); Assert.Equal(1, Coding007.FibonacciSum(1)); Assert.Equal(1, Coding007.FibonacciSum2(1)); } [Fact] public void Test2() { Assert.Equal(1, Coding007.Fibonacci(2)); Assert.Equal(1, Coding007.Fibonacci2(2)); Assert.Equal(2, Coding007.FibonacciSum(2)); Assert.Equal(2, Coding007.FibonacciSum2(2)); } [Fact] public void Test3() { Assert.Equal(2, Coding007.Fibonacci(3)); Assert.Equal(2, Coding007.Fibonacci2(3)); Assert.Equal(4, Coding007.FibonacciSum(3)); Assert.Equal(4, Coding007.FibonacciSum2(3)); } [Fact] public void Test4() { Assert.Equal(3, Coding007.Fibonacci(4)); Assert.Equal(3, Coding007.Fibonacci2(4)); Assert.Equal(7, Coding007.FibonacciSum(4)); Assert.Equal(7, Coding007.FibonacciSum2(4)); } [Fact] public void Test5() { Assert.Equal(5, Coding007.Fibonacci(5)); Assert.Equal(5, Coding007.Fibonacci2(5)); Assert.Equal(12, Coding007.FibonacciSum(5)); Assert.Equal(12, Coding007.FibonacciSum2(5)); } [Fact] public void Test6() { Assert.Equal(8, Coding007.Fibonacci(6)); Assert.Equal(8, Coding007.Fibonacci2(6)); Assert.Equal(20, Coding007.FibonacciSum(6)); Assert.Equal(20, Coding007.FibonacciSum2(6)); }
想入非非:扩展思维,发挥想象
1. 熟悉递归
2. 熟悉斐波那契数列
3. 斐波那契数列求和
4. 知道有公式的就用公式,不要自己去循环就算,就像1+2+3+......,用高斯定理直接算结果,不要再循环了