• 《剑指offer》第五十九题(滑动窗口的最大值)


    // 面试题59(一):滑动窗口的最大值
    // 题目:给定一个数组和滑动窗口的大小,请找出所有滑动窗口里的最大值。例如,
    // 如果输入数组{2, 3, 4, 2, 6, 2, 5, 1}及滑动窗口的大小3,那么一共存在6个
    // 滑动窗口,它们的最大值分别为{4, 4, 6, 6, 6, 5},
    
    #include <iostream>
    #include <vector>
    #include <deque>
    
    using namespace std;
    
    vector<int> maxInWindows(const vector<int>& num, unsigned int size)
    {
        vector<int> maxInWindows;
        if (num.size() >= size && size >= 1)
        {
            deque<int> index;
    
            for (unsigned int i = 0; i < size; ++i)//先找出第一个滑窗里的最大值的坐标,放入队列头
            {
                while (!index.empty() && num[i] >= num[index.back()])
                    index.pop_back();
    
                index.push_back(i);
            }
    
            for (unsigned int i = size; i < num.size(); ++i)//对之后的每个点
            {
                maxInWindows.push_back(num[index.front()]);//先把队列头放入maxInWindows
    
                while (!index.empty() && num[i] >= num[index.back()])//如果队列不为空且当前点大于尾节点,就删除尾节点,直到尾节点不小于当前节点
                    index.pop_back();
                if (!index.empty() && index.front() <= (int)(i - size))//如果队列不为空,且头结点的位置小于当前位置减去滑窗长度,说明这个最大值不再滑窗内了,删除
                    index.pop_front();
    
                index.push_back(i);//插入当前点的坐标
            }
            maxInWindows.push_back(num[index.front()]);//最后那个值没在循环中插入maxInWindows,要额外写一句
        }
    
        return maxInWindows;
    }
    
    // ====================测试代码====================
    void Test(const char* testName, const vector<int>& num, unsigned int size, const vector<int>& expected)
    {
        if (testName != nullptr)
            printf("%s begins: ", testName);
    
        vector<int> result = maxInWindows(num, size);
    
        vector<int>::const_iterator iterResult = result.begin();
        vector<int>::const_iterator iterExpected = expected.begin();
        while (iterResult < result.end() && iterExpected < expected.end())
        {
            if (*iterResult != *iterExpected)
                break;
    
            ++iterResult;
            ++iterExpected;
        }
    
        if (iterResult == result.end() && iterExpected == expected.end())
            printf("Passed.
    ");
        else
            printf("FAILED.
    ");
    }
    
    void Test1()
    {
        int num[] = { 2, 3, 4, 2, 6, 2, 5, 1 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 4, 4, 6, 6, 6, 5 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 3;
    
        Test("Test1", vecNumbers, size, vecExpected);
    }
    
    void Test2()
    {
        int num[] = { 1, 3, -1, -3, 5, 3, 6, 7 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 3, 3, 5, 5, 6, 7 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 3;
    
        Test("Test2", vecNumbers, size, vecExpected);
    }
    
    // 输入数组单调递增
    void Test3()
    {
        int num[] = { 1, 3, 5, 7, 9, 11, 13, 15 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 7, 9, 11, 13, 15 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 4;
    
        Test("Test3", vecNumbers, size, vecExpected);
    }
    
    // 输入数组单调递减
    void Test4()
    {
        int num[] = { 16, 14, 12, 10, 8, 6, 4 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 16, 14, 12 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 5;
    
        Test("Test4", vecNumbers, size, vecExpected);
    }
    
    // 滑动窗口的大小为1
    void Test5()
    {
        int num[] = { 10, 14, 12, 11 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 10, 14, 12, 11 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 1;
    
        Test("Test5", vecNumbers, size, vecExpected);
    }
    
    // 滑动窗口的大小等于数组的长度
    void Test6()
    {
        int num[] = { 10, 14, 12, 11 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        int expected[] = { 14 };
        vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
    
        unsigned int size = 4;
    
        Test("Test6", vecNumbers, size, vecExpected);
    }
    
    // 滑动窗口的大小为0
    void Test7()
    {
        int num[] = { 10, 14, 12, 11 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        vector<int> vecExpected;
    
        unsigned int size = 0;
    
        Test("Test7", vecNumbers, size, vecExpected);
    }
    
    // 滑动窗口的大小大于输入数组的长度
    void Test8()
    {
        int num[] = { 10, 14, 12, 11 };
        vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
    
        vector<int> vecExpected;
    
        unsigned int size = 5;
    
        Test("Test8", vecNumbers, size, vecExpected);
    }
    
    // 输入数组为空
    void Test9()
    {
        vector<int> vecNumbers;
        vector<int> vecExpected;
    
        unsigned int size = 5;
    
        Test("Test9", vecNumbers, size, vecExpected);
    }
    
    int main(int argc, char* argv[])
    {
        Test1();
        Test2();
        Test3();
        Test4();
        Test5();
        Test6();
        Test7();
        Test8();
        Test9();
        system("pause");
        return 0;
    }
  • 相关阅读:
    图像维纳滤波实现(2)
    C++之路 变量和基本类型(2)
    [C# 线程处理系列]专题四:线程同步
    [转]排列组合
    Bessie Come Home(!)先存着
    The ONE v1.4.1 Readme
    myeclipse字体修改
    vm 三种上网方式
    Cow Tours
    C++中, 构造函数和析构函数能不能被显示调用?
  • 原文地址:https://www.cnblogs.com/CJT-blog/p/10545627.html
Copyright © 2020-2023  润新知