【4】在矩阵中查找是否存在某个数

在二维数组中查找 某一元素

1、题目：在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按
照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个
整数，判断数组中是否含有该整数。

2、分析：选定数组中右上角或者（左下角）元素，与待查找的数字进行比较，如果待查找数字大于右上角元素，行数加1，继续比较；如果待查找数字小于右上角元素，列数减一再比较，这样每次比较完，都会少掉一行或者一列，时间复杂度为O(n);

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int Find(int*matrix,int rows,int cols,int findkey)
{
    int found = -1;
    if(matrix == NULL ||rows < 0 || cols < 0)
    {
        return found;
    }
    if(matrix != NULL ||rows > 0 || cols > 0)
    {
        int row = 0;
        int col = cols - 1;
        while(col >= 0 && row < rows)
        {
            if(matrix[row*cols+col] == findkey)
            {
                found = 1;
                break;
            }
            else
            {
                if(matrix[row*cols+col] > findkey)
                {
                    --col;
                }
                else ++row;
            }
        }
    }
    return found;
}



//===================测试代码=================
void test(int* matrix,int rows,int cols,int findkey)
{
    if(matrix == NULL || rows <= 0 ||cols<= 0)
    {
        printf("参数出现传送错误\n");
        return ;
    }
    int ret = 0;
    ret = Find(matrix, M, N, findkey);
        if(ret == -1)
        {
            printf("查找的数字在本数组中不存在\n");
        }
        else
        printf("存在\n");


}
//================存在于数组中的三种情况=====================
//1.数组中存在待查找的数字min< 10<max
void test1()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 10;
    printf("test1()中情况：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}
//2.数组中存在待查找的数字min
void test2()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 1;
    printf("\ntest2()中情况：：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}
//3.数组中存在待查找的数字max
void test3()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 15;
    printf("\ntest3()中情况：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}
//===============================不存在的三种情况===================
//4.数组中不存在待查找的数字，findkey<min
void test4()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = -1;
    printf("\ntest4()中情况：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}
//5.数组中不存在待查找的数字，findkey>max
void test5()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 16;
    printf("\ntest5()中情况：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}
//6.数组中不存在待查找的数字，min<findkey<max
void test6()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 5;
    printf("\ntest6()中情况：findkey= %d ",findkey);
    test((int* )arr,4,4,findkey);
}

//=========================测试代码的鲁棒性================
void test7()
{
    int arr[][4] = {{1,2,8,9}, 
                    {2,4,9,12},
                    {4,7,10,13},
                    {6,8,11,15}};
    int findkey = 1;
    printf("\ntest7()中情况：");
    test(NULL,4,4,findkey);
}
int main()
{
    test1();
    test2();
    test3();
    test4();
    test5();
    test6();
    test7();
    return 0;
}