zoukankan      html  css  js  c++  java
  • 递归与循环

    1. 例子:

    #include <iostream>
    using namespace std;
    
    //循环实现累加:
    int Sum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
        {
            sum += i;
        }
        return sum;
    }
    //递归实现累加:
    int RecursiveSum(int n)
    {
        if (n == 0)
        { 
            return 0;
    
        } 
        else
        {
            return RecursiveSum(n - 1) + n;
        }
    }
    
    void main()
    {
        int num;
        cout << "请输入要累加的数:";
        cin >> num;
    
        int sum1 = Sum(num);
        int sum2 = RecursiveSum(num);
    
        cout << "循环结果为:" << sum1 << endl;
        cout << "递归结果为:" << sum2 << endl;
    
        system("pause");
    }

        

    2. 递归转非递归:

    #include <iostream>
    #include <stack>
    
    using namespace std;
    
    //递归:
    void Print1(int n) //打印1~n的数(顺序)
    {
        if (n <= 0)
        {
            return;
        } 
        else
        {
            Print1(n - 1);        //交换这两行位置,则为逆序
            cout << n << "  " ;
        }
    }
    
    //非递归:
    void RecursivePrint1(int n)
    {
        stack<int> mystack;                        /* 1.初始化栈 */
    
    L0:                                            /* 2.设入口标号 */
        if (n <= 0)                                /* 3.非递归调用和返回部分照搬 */
        {
            while (!mystack.empty())            /* 5.返回部分替换 */
            {
                cout << mystack.top() << "  " ;
                mystack.pop();
            }
    
            return ;
        }
        else                                    /* 4.递归调用Print1(n - 1),替换 */
        {
            mystack.push(n);
            n = n - 1;
    
            goto L0;
        }
    }
    
    //递归(逆序):
    void Print2(int n) //打印1~n的数(顺序)
    {
        if (n <= 0)
        {
            return;
        }
        else
        {
            cout << n << "  ";
            Print2(n - 1);        
        }
    }
    //非递归:
    void RecursivePrint2(int n)
    {
        stack<int> mystack;                            /* 1.初始化栈 */
    
    L0:                                                /* 2.设入口标号 */
        if (n <= 0)                                    /* 3.非递归调用和返回部分照搬 */
        {
            return;
        }
        else                                        /* 4.递归调用Print2(n - 1),替换 */
        {
            mystack.push(n);    //吃一个,吐一个,逆序
    
            while (!mystack.empty())                /* 5.返回部分替换 */
            {
                cout << mystack.top() << "  ";
                mystack.pop();
            }
    
            n = n - 1;
    
            goto L0;
        }
    }
    
    void main()
    {
        cout << "递归(顺序):" << endl;
        Print1(10);
        cout << "
    
    非递归(顺序):" << endl;
        RecursivePrint1(10);
            
    
        cout << "
    
    递归(逆序):" << endl;
        Print2(10);
        cout << "
    
    非递归(逆序):" << endl;
        RecursivePrint2(10);
    
        cin.get();
    }

        

    3. 斐波那契数列 递归转非递归:

    #include <iostream>
    #include <stack>
    
    using namespace std;
    
    //递归斐波那契:
    int Fibonacci(int n)
    {
        if (n == 1)
        {
            return 1;
        }
        else if (n == 2)
        {
            return 1;
        }
        else
        {
            return Fibonacci(n - 1) + Fibonacci(n - 2);//树状递归
        }
    }
    
    //(方法一)非递归斐波那契:
    int RecursiveFibonacci1(int n)
    {
        if (n==1)
        {
            return 1;
        }
        else if (n == 2)
        {
            return 1;
        }
    
        int f1, f2, f3;
        f1 = f2 = 1;
        for (int i = 2; i < n; i++)
        {
            f3 = f1 + f2;
            f1 = f2;
            f2 = f3;    //轮替
        }
    
        return f3;
    }
    //(方法二)非递归斐波那契:
    int RecursiveFibonacci2(int n)
    {
        if (n == 1)
        {
            return 1;
        }
        else if (n == 2)
        {
            return 1;
        }
    
        stack<int> mystack;
    
        int f1, f2, f3;
        f1 = f2 = 1;
        int i = 2;
    
    L0:
        if (i < n)
        {
            mystack.push(f1);
            mystack.push(f2);
            f3 = 0;
            while (!mystack.empty())
            {
                f3 = f3 + mystack.top();
                mystack.pop();
            }
            f1 = f2;
            f2 = f3;
            i++;
    
            goto L0;
        }
    
        return f3;
    }
    
    void main()
    {
        int num;
        cout << "斐波那契个数:";
        cin >> num;
    
        cout << "
    
    递归斐波那契:" << endl;
        for (int i = 1; i <= num; i++)
        {
            cout << Fibonacci(i) << "  ";
        }
    
        cout << "
    
    (方法一)非递归斐波那契:" << endl;
        for (int i = 1; i <= num; i++)
        {
            cout << RecursiveFibonacci1(i) << "  ";
        }
    
        cout << "
    
    (方法二)非递归斐波那契:" << endl;
        for (int i = 1; i <= num; i++)
        {
            cout << RecursiveFibonacci2(i) << "  ";
        }
    
    
        cout <<"
    
    ";
        system("pause");
    }

        

  • 相关阅读:
    hdu4622(后缀自动机模板)
    51nod1445(最短路)
    Java原子类--框架
    Java锁--Semaphore
    Java锁--CyclicBarrier
    Java锁--CountDownLatch
    Java锁--共享锁和ReentrantReadWriteLock
    Java锁--LockSupport
    Java锁--Condition
    Java锁--非公平锁
  • 原文地址:https://www.cnblogs.com/si-lei/p/9551074.html
Copyright © 2011-2022 走看看