【动态规划专题】2：矩阵的最小路径和

《程序员代码面试指南--IT名企算法与数据结构题目最优解》 左程云 著

矩阵的最小路径和

【题目】
给定一个矩阵m，从左上角开始每次只能向右或者向下走，最后到达右下角位置，路径上所有的数字累加起来就是路径和，返回所有路径中最小的路径和。

【举例】
如果给定的m如下：
1 3 5 9
8 1 3 4
5 0 6 1
8 8 4 0
路径1，3，1,0,6,1,0是所有路径中路径和最小的，所以返回12.


关键思路：
假设有一个M*N的数组 dp[M][N], dp[i][j]的值表示从左上角（0,0）位置走到（i,j）位置的最小路径和。
第0行和第0列比较特殊，是不断累加的结果
1 4 9 18
9,
14
22

其他行和列，dp[i][j] = Min(dp[i-1][j], dp[i][j-1]) + m[i][j];

所以，我们可以从最右下角，倒推到左上角。

从下到上思考，从上到下解决。

#include <iostream>
#include <stack>
#include <exception>
using namespace std;

int minOfTwoNum(int A, int B)
{
    return A > B ? B : A;
}

int MaxOfTwoNum(int A, int B)
{
    return A > B ? A : B;
}
//申请一个M*N的二位数组空间
int MinPathRoad(int* arrPath, const int rows,const  int cols)
{
    if (arrPath == nullptr || rows < 0 || cols < 0)
    {
        throw new std::exception("Invalid para");
    }
    if (rows == 0 || cols == 0)
    {
        return 0;
    }
    int result = 0;
    int** copyPath =new int*[rows];
    for (int i = 0; i < rows; i++)
    {
        copyPath[i] = new int[cols];
    }


    ///这里有可优化空间。能用一维数组解决的，不要用二维数组
    //for (int i = 0; i < rows; i++)
    //{
    //    for (int j = 0; j < cols; j++)
    //    {
    //        if (i == 0 && j == 0)
    //        {
    //            copyPath[i][j] = *(arrPath + i*cols + j);
    //        }
    //        if (i == 0 && j>0)
    //        {
    //            copyPath[i][j] = *(arrPath + i*cols + j) + *(arrPath + i*cols + j - 1);
    //        }
    //        if (j == 0 && i > 0)
    //        {
    //            copyPath[i][j] = *(arrPath + i*cols + j) + *(arrPath + (i - 1)*cols + j);
    //        }
    //    }
    //}

    copyPath[0][0] = *(arrPath);

    for (int j = 1; j < cols; j++)
    {
        copyPath[0][j] = *(arrPath + j) + copyPath[0][j-1]; ///这里容易出错
    }

    for (int i = 1; i < rows; i++)
    {
        copyPath[i][0] = *(arrPath + i*cols) + copyPath[i - 1][0]; ///这里容易出错
    }

    ////时间复杂度是O(M*N)
    for (int i = 1; i < rows; i++)
    {
        for (int j = 1; j < cols; j++)
        {
            int tempValue = minOfTwoNum(copyPath[i - 1][j], copyPath[i][j - 1]);
            copyPath[i][j] = minOfTwoNum(copyPath[i - 1][j], copyPath[i][j - 1]) + *(arrPath + i*cols + j);
        }
    }

    result = copyPath[rows - 1][cols - 1];


    for (int i = 0; i < rows; i++)
    {
        delete[] copyPath[i];
    }
    delete[] copyPath;


    return result;
}
//申请一个一维数组空间
int MinPathRoad2(int* arrPath, const int rows, const  int cols)
{
    if (arrPath == nullptr || rows < 0 || cols < 0)
    {
        throw new std::exception("Invalid para");
    }
    if (rows == 0 || cols == 0)
    {
        return 0;
    }
    int result = 0;
    //用一个一维数组来存储临时变量
    int* copyPath = new int[cols];

    for (int j = 0; j < cols; j++)
    {
        if (j == 0)
        {
            copyPath[j] = *(arrPath  + j);
        }
        if (j>0)
        {
            copyPath[j] = *(arrPath + 0 * cols + j) + copyPath[j-1];
        }
    }

    for (int i = 1; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
        {
            if (j == 0)
            {
                copyPath[j] = copyPath[j] + *(arrPath + i*cols + j);
            }
            else
            {
                //int tempValue = minOfTwoNum(copyPath[j], copyPath[j - 1]);
                copyPath[j] = minOfTwoNum(copyPath[j], copyPath[j - 1]) + *(arrPath + i*cols + j);
            }

        }
    }

    result = copyPath[cols - 1];


    delete[] copyPath;

    return result;
}
//申请一个一维数组空间，且取rows,cols更小的那个
int MinPathRoad3(int* arrPath, const int rows, const  int cols)
{    
    //=============步子不能迈太大，先从if,else入手。然后再去整合到一起。
    if (arrPath == nullptr || rows < 0 || cols < 0)
    {
        throw new std::exception("Invalid para");
    }
    if (rows == 0 || cols == 0)
    {
        return 0;
    }

    ////分配一行，cols个元素，按行扫描
    if (rows >= cols)
    {
        int result = 0;
        //用一个一维数组来存储临时变量
        int* copyPath = new int[cols];

        for (int j = 0; j < cols; j++)
        {
            if (j == 0)
            {
                copyPath[j] = *(arrPath + j);
            }
            if (j>0)
            {
                copyPath[j] = *(arrPath + 0 * cols + j) + copyPath[j - 1];
            }
        }

        for (int i = 1; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                if (j == 0)
                {
                    copyPath[j] = copyPath[j] + *(arrPath + i*cols + j);
                }
                else
                {
                    //int tempValue = minOfTwoNum(copyPath[j], copyPath[j - 1]);
                    copyPath[j] = minOfTwoNum(copyPath[j], copyPath[j - 1]) + *(arrPath + i*cols + j);
                }

            }
        }

        result = copyPath[cols - 1];

        delete[] copyPath; //不要删太早了

        return result;
    }
    else
    {
        ////分配一列，rows个元素，按列扫描
        int result = 0;
        //用一个一维数组来存储临时变量
        int* copyPath = new int[rows];

        for (int i = 0; i < rows; i++)
        {
            if (i == 0)
            {
                copyPath[i] = *(arrPath + i);
            }
            if (i>0)
            {
                copyPath[i] = *(arrPath + i * cols ) + copyPath[i - 1];
            }
        }

        for (int j = 1; j < cols; j++)
        {
            for (int i = 0; i < rows; i++)
            {
                if (i == 0)
                {
                    copyPath[i] = copyPath[i] + *(arrPath + i*cols + j);
                }
                else
                {
                    copyPath[i] = minOfTwoNum(copyPath[i], copyPath[i - 1]) + *(arrPath + i*cols + j);
                }

            }
        }

        result = copyPath[rows - 1];
        delete[] copyPath; //不要删太早了
        return result;
    }
}
//申请一个一维数组空间，且取rows,cols更小的那个。不使用if,else，而是整合rows,cols判断
int MinPathRoad4(int* arrPath, const int rows, const  int cols)
{
    if (arrPath == nullptr || rows < 0 || cols < 0)
    {
        throw new std::exception("Invalid para");
    }
    if (rows == 0 || cols == 0)
    {
        return 0;
    }

    bool bRowMore = (rows>cols);//rows是否更大
    int result = 0;

    //用一个一维数组来存储临时变量
    int less = minOfTwoNum(rows, cols);
    int more = MaxOfTwoNum(rows, cols);
    int* copyPath = new int[less];

    copyPath[0] = *(arrPath);
    for (int i = 1; i < less; i++)
    {
        //bRowMore  ? arr[j][0] : arr[0][j]   
        copyPath[i] = copyPath[i - 1] + (bRowMore ? *(arrPath + i) : *(arrPath + i * cols));
    }//


    for (int i = 1; i < more; i++)
    {
        for (int j = 0; j < less; j++)
        {
            if (j == 0)
            {
                copyPath[0] = copyPath[0] + (bRowMore ? *(arrPath + i*cols+j) : *(arrPath + i));
            }
            else
            {
                int tempValue = minOfTwoNum(copyPath[j], copyPath[j - 1]);
                copyPath[j] = minOfTwoNum(copyPath[j], copyPath[j - 1]) + (bRowMore ? *(arrPath + i*cols + j) : *(arrPath + i*rows + j));
            }
        }
    }
    result = copyPath[less - 1];

    delete[] copyPath;
    return result;

}

一定不能省掉测试用例。

用足够丰富的测试用例，去验证代码功能的完整性。

//====================测试用例=================
// 1 3 5 9 
// 8 1 3 4  
// 5 0 6 1
// 8 8 4 0
void test1()
{
    cout << "Test1================expected:12" << endl;
    int arrPath[4][4] = { { 1, 3, 5, 9 }, { 8, 1, 3, 4 }, { 5, 0, 6, 1 }, { 8, 8, 4, 0 } };
    cout << MinPathRoad((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 4) << endl;
}


//
// 1 1 1
// 3 4 2
// 7 8 3
// 6 6 0
void test2()
{
    cout << "Test2================expected:8" << endl;
    int arrPath[4][3] = {{ 1,1,1 }, { 3,4,2 }, { 7,8,3 }, {6,6,0 } };
    cout << MinPathRoad((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 3) << endl;
}

//
// 1 1 2
// 1 4 2
// 2 8 3
// 3 1 0
void test3()
{
    cout << "Test3================expected:8" << endl;
    int arrPath[4][3] = { { 1, 1, 2 }, { 1, 4, 2 }, {2, 8, 3 }, { 3, 1, 0 } };
    cout << MinPathRoad((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 3) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 3) << endl;
}


//
// 1  2   3   4
// 5  0   7   8
// 9  0  11  12
// 0  14  0  0
void test4()
{
    cout << "Test4================expected:14" << endl;
    int arrPath[4][4] = { { 1, 2, 3,4 }, { 5,0,7,8}, { 9,0,11,12 }, {0,14, 0,0 } };
    cout << MinPathRoad((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 4) << endl;
}

//
// 1 1 1 1
// 1 1 1 1
// 1 1 1 1
// 1 1 1 1
void test5()
{
    cout << "Test5================expected:7" << endl;
    int arrPath[4][4] = { { 1, 1, 1, 1 }, { 1, 1, 1, 1 }, { 1, 1, 1, 1 }, { 1, 1, 1, 1 } };
    cout << MinPathRoad((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 4) << endl;
}


//
// 0 0 0 0
// 0 1 1 0
// 0 0 0 1
// 0 1 0 0
void test6()
{
    cout << "Test6================expected:0" << endl;
    int arrPath[4][4] = { { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 } };
    cout << MinPathRoad((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad2((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad3((int *)arrPath, 4, 4) << endl;
    cout << MinPathRoad4((int *)arrPath, 4, 4) << endl;
}


int main()
{
    test1();
    test2();
    test3();
    test4();
    test5();
    test6();

    system("pause");
    return 0;
}
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原文地址：https://www.cnblogs.com/music-liang/p/12054353.html