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  • 《剑指offer》第六十一题:扑克牌的顺子

    // 面试题61:扑克牌的顺子
    // 题目:从扑克牌中随机抽5张牌,判断是不是一个顺子,即这5张牌是不是连续的。
    // 2~10为数字本身,A为1,J为11,Q为12,K为13,而大、小王可以看成任意数字。
    
    #include <cstdio>
    #include <cstdlib>
    
    int Compare(const void* arg1, const void* arg2);
    
    bool IsContinuous(int* numbers, int length)
    {
        if (numbers == nullptr || length < 0)
            return false;
    
        qsort(numbers, length, sizeof(int), Compare);
    
        int numberOfZero = 0;  //大小王个数
        for (int i = 0; i < length && numbers[i] == 0; i++)
            ++numberOfZero;  //忘了for循环可以多个条件
    
    
        int numberOfGap = 0;
        for (int i = numberOfZero; i < length - 1; i++)  //从第一个不是0的数字开始
        {
            if (numbers[i + 1] == numbers[i])  //如果出现对子, 则不可能是顺子
                return false;
    
            numberOfGap += (numbers[i + 1] - numbers[i] - 1);  //差2 == 间隙1
        }
    
        return (numberOfZero >= numberOfGap) ? true : false;
    }
    
    int Compare(const void* arg1, const void* arg2)
    {
        return *(int*) arg1 - *(int*) arg2;
    }
    // ====================测试代码====================
    void Test(const char* testName, int* numbers, int length, bool expected)
    {
        if (testName != nullptr)
            printf("%s begins: ", testName);
    
        if (IsContinuous(numbers, length) == expected)
            printf("Passed.
    ");
        else
            printf("Failed.
    ");
    }
    
    void Test1()
    {
        int numbers[] = { 1, 3, 2, 5, 4 };
        Test("Test1", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    void Test2()
    {
        int numbers[] = { 1, 3, 2, 6, 4 };
        Test("Test2", numbers, sizeof(numbers) / sizeof(int), false);
    }
    
    void Test3()
    {
        int numbers[] = { 0, 3, 2, 6, 4 };
        Test("Test3", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    void Test4()
    {
        int numbers[] = { 0, 3, 1, 6, 4 };
        Test("Test4", numbers, sizeof(numbers) / sizeof(int), false);
    }
    
    void Test5()
    {
        int numbers[] = { 1, 3, 0, 5, 0 };
        Test("Test5", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    void Test6()
    {
        int numbers[] = { 1, 3, 0, 7, 0 };
        Test("Test6", numbers, sizeof(numbers) / sizeof(int), false);
    }
    
    void Test7()
    {
        int numbers[] = { 1, 0, 0, 5, 0 };
        Test("Test7", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    void Test8()
    {
        int numbers[] = { 1, 0, 0, 7, 0 };
        Test("Test8", numbers, sizeof(numbers) / sizeof(int), false);
    }
    
    void Test9()
    {
        int numbers[] = { 3, 0, 0, 0, 0 };
        Test("Test9", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    void Test10()
    {
        int numbers[] = { 0, 0, 0, 0, 0 };
        Test("Test10", numbers, sizeof(numbers) / sizeof(int), true);
    }
    
    // 有对子
    void Test11()
    {
        int numbers[] = { 1, 0, 0, 1, 0 };
        Test("Test11", numbers, sizeof(numbers) / sizeof(int), false);
    }
    
    // 鲁棒性测试
    void Test12()
    {
        Test("Test12", nullptr, 0, false);
    }
    
    int main(int argc, char* argv[])
    {
        Test1();
        Test2();
        Test3();
        Test4();
        Test5();
        Test6();
        Test7();
        Test8();
        Test9();
        Test10();
        Test11();
        Test12();
    
        return 0;
    }
    测试代码

    分析:建模以后真简单。

    class Solution {
    public:
        bool IsContinuous( vector<int> numbers ) {
            
            int length = numbers.size();
            
            if (length < 1)
                return false;
            
            sort(numbers.begin(), numbers.end());
            
            int numberOfZero = 0;
            for (int i = 0; i < length && numbers[i] == 0; ++i)
                ++numberOfZero;
            
            int numberOfGap = 0;
            for (int i = numberOfZero; i < length - 1; ++i)
            {
                if (numbers[i + 1] == numbers[i])
                    return false;
                
                numberOfGap += (numbers[i + 1] - numbers[i] - 1);
            }
            return (numberOfZero >= numberOfGap)? true:false;
        }
    };
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  • 原文地址:https://www.cnblogs.com/ZSY-blog/p/12686670.html
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