• 剑指offer——面试题14:剪绳子


    // 面试题14:剪绳子
    // 题目:给你一根长度为n绳子,请把绳子剪成m段(m、n都是整数,n>1并且m≥1)。
    // 每段的绳子的长度记为k[0]、k[1]、……、k[m]。k[0]*k[1]*…*k[m]可能的最大乘
    // 积是多少?例如当绳子的长度是8时,我们把它剪成长度分别为2、3、3的三段,此
    // 时得到最大的乘积18。
    
    #include <iostream>
    #include <cmath>
    
    // ====================动态规划====================
    int maxProductAfterCutting_solution1(int length)
    {
        if(length < 2)
            return 0;
        if(length == 2)
            return 1;
        if(length == 3)
            return 2;
    
        int* products = new int[length + 1];
        products[0] = 0;
        products[1] = 1;
        products[2] = 2;
        products[3] = 3;
    
        int max = 0;
        for(int i = 4; i <= length; ++i)
        {
            max = 0;
            for(int j = 1; j <= i / 2; ++j)
            {
                int product = products[j] * products[i - j];
                if(max < product)
                    max = product;
    
                products[i] = max;
            }
        }
    
        max = products[length];
        delete[] products;
    
        return max;
    }
    
    // ====================贪婪算法====================
    int maxProductAfterCutting_solution2(int length)
    {
        if(length < 2)
            return 0;
        if(length == 2)
            return 1;
        if(length == 3)
            return 2;
    
        // 尽可能多地减去长度为3的绳子段
        int timesOf3 = length / 3;
    
        // 当绳子最后剩下的长度为4的时候,不能再剪去长度为3的绳子段。
        // 此时更好的方法是把绳子剪成长度为2的两段,因为2*2 > 3*1。
        if(length - timesOf3 * 3 == 1)
            timesOf3 -= 1;
    
        int timesOf2 = (length - timesOf3 * 3) / 2;
    
        return (int) (pow(3, timesOf3)) * (int) (pow(2, timesOf2));
    }
    
    // ====================测试代码====================
    void test(const char* testName, int length, int expected)
    {
        int result1 = maxProductAfterCutting_solution1(length);
        if(result1 == expected)
            std::cout << "Solution1 for " << testName << " passed." << std::endl;
        else
            std::cout << "Solution1 for " << testName << " FAILED." << std::endl;
    
        int result2 = maxProductAfterCutting_solution2(length);
        if(result2 == expected)
            std::cout << "Solution2 for " << testName << " passed." << std::endl;
        else
            std::cout << "Solution2 for " << testName << " FAILED." << std::endl;
    }
    
    void test1()
    {
        int length = 1;
        int expected = 0;
        test("test1", length, expected);
    }
    
    void test2()
    {
        int length = 2;
        int expected = 1;
        test("test2", length, expected);
    }
    
    void test3()
    {
        int length = 3;
        int expected = 2;
        test("test3", length, expected);
    }
    
    void test4()
    {
        int length = 4;
        int expected = 4;
        test("test4", length, expected);
    }
    
    void test5()
    {
        int length = 5;
        int expected = 6;
        test("test5", length, expected);
    }
    
    void test6()
    {
        int length = 6;
        int expected = 9;
        test("test6", length, expected);
    }
    
    void test7()
    {
        int length = 7;
        int expected = 12;
        test("test7", length, expected);
    }
    
    void test8()
    {
        int length = 8;
        int expected = 18;
        test("test8", length, expected);
    }
    
    void test9()
    {
        int length = 9;
        int expected = 27;
        test("test9", length, expected);
    }
    
    void test10()
    {
        int length = 10;
        int expected = 36;
        test("test10", length, expected);
    }
    
    void test11()
    {
        int length = 50;
        int expected = 86093442;
        test("test11", length, expected);
    }
    
    int main(int agrc, char* argv[])
    {
        test1();
        test2();
        test3();
        test4();
        test5();
        test6();
        test7();
        test8();
        test9();
        test10();
        test11();
    
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/acm-jing/p/10392468.html
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