行、列递增的二维数组数字查找

二维数组四分查找

问题：
一个二维数组，但行递增、单列递增，编写程序，在数组中查找某个数字（key），要求时间复杂度小于O（N）。
示例数组
1 3 4
2 4 5
4 5 6

分析：
对这个问题，首先想到的就是二分查找，对一位数组效率最高的查找方法，能不能用类似的方法来处理这个问题呢？
先看一下二分查找

根据mid与key的关系确定下次查找的范围或输出结束
那么二维数组呢
_ 绿色表示比较的数字位置
_ 浅灰色区域表示该区域小于key
_ 深灰色区域表示该区域大于key
_ 浅蓝色表示key可能出现的区域
_ 深蓝色表示下次搜索的区域

中心值大于key的情况与小于时类似

初始代码
（注释见优化后的代码）

#include<stdio.h>void act_search(int a[][3], int key, int left, int right, int top, int bot){    if ((left <= right) && (top <= bot))    {        int mid_row = (top + bot) >> 1;        int mid_col = (left + right) >> 1;        if (a[mid_row][mid_col] != key)        {            if (a[mid_row][mid_col] > key)            {                if (a[top][mid_col] >= key)                {                    act_search(a, key, left, mid_col - 1, top, mid_row - 1);                    if (a[top][mid_col] == key)                    {                        printf("a[%d][%d] = %d\n", top, mid_col, a[top][mid_col]);                    }                }                else//a[top][mid_col] < key                {                    act_search(a, key, mid_col, right, top, mid_row - 1);                }                if (a[mid_row][left] >= key)                {                    if (a[top][mid_col] <= key)                    {                        act_search(a, key, left, mid_col - 1, top, mid_row - 1);                    }                    if (a[mid_row][left] == key)                    {                        printf("a[%d][%d] = %d\n", top, mid_col, a[top][mid_col]);                    }                }                else//a[mid_row][left]<key                {                    act_search(a, key, left, mid_col - 1, mid_row, bot);                }            }            else    //(a[mid_row][mid_col] < key)            {                if (a[mid_row][right] <= key)                {                    act_search(a, key, mid_col + 1, right, mid_row + 1, bot);                    if (a[mid_row][bot] == key)                    {                        printf("a[%d][%d] = %d\n", mid_col, right, a[mid_col][right]);                    }                }                else   //a[mid_row][bot] > key                {                    act_search(a, key, mid_col + 1, right, top, mid_row);                }                if (a[bot][mid_col] <= key)                {                    if (a[mid_row][right] > key)                    {                        act_search(a, key, mid_col + 1, right, mid_row + 1, bot);                    }                    if (a[bot][mid_col] == key)                    {                        printf("a[%d][%d] = %d\n", bot, mid_col, a[bot][mid_col]);                    }                }                else  //a[bot][mid_col] > key                {                    act_search(a, key, left, mid_col, mid_row + 1, bot);                }            }        }        else     //a[mid_row][mid_col] != key        {            printf("a[%d][%d] = %d\n", mid_row, mid_col, a[mid_row][mid_col]);            act_search(a, key, left, mid_col - 1, mid_row + 1, bot);            act_search(a, key, mid_col + 1, right, top, mid_row - 1);        }    }}void search(int a[][3], int key, int rol, int cow){    int left = 0;    int right = cow - 1;    int top = 0;    int bot = rol - 1;    act_search(a, key, left, right, top, bot);}int main(){    int a[][3]={{ 1, 3, 4 },{2, 4, 5},{4, 5, 6}};    int rol = sizeof(a) / sizeof(a[0]);    int cow = sizeof(a[0]) / sizeof(a[0][0]);    int key = 4;    search(a, key, rol, cow);    return 0;}

优化
有一个很大的问题

在函数内部写死了列数让函数局限性变得非常大
而事实上我们知道，二维数组在内存中的存储是连续的，二维数组可以用一维数组来表示和处理，p为首元素指针，COL为二维数组列数，则a[row][col]等价于p[row*COL+col]，那么search函数只需接收数组首元素地址而不需接收整个数组，act_search也只需接收首元素地址，另外加列数就好了。然后再将二维数组一维表示就OK了。

#include<stdio.h>void act_search(int *a, int key, int left, int right, int top, int bot, int col){    if ((left <= right) && (top <= bot))  //判别条件，左值不大于右值，顶值不大于底值    {        int mid_row = (top + bot) >> 1;  //列中心        int mid_col = (left + right) >> 1;  //行中心        if (a[mid_row*col+mid_col] != key)  //数组中心不等于key        {            if (a[mid_row*col + mid_col] > key)  // 数组中心大于key，排除右下区            {                if (a[top*col + mid_col] >= key)    //顶行中心大于等于key，排除右上区域，剩余左上区                {                    act_search(a, key, left, mid_col - 1, top, mid_row - 1, col);  搜索左上区                    if (a[top*col + mid_col] == key)    //找到key则输出                    {                        printf("a[%d][%d] = %d\n", top, mid_col, a[top*col + mid_col]);                    }                }                else   //a[top][mid_col] < key  //顶行中心小于key，搜索包括所在列的右上区                {                    act_search(a, key, mid_col, right, top, mid_row - 1, col);                }                if (a[mid_row*col + left] >= key)   //左边中心不小于key，排除左下区                {                    if (a[top*col + mid_col] <= key)   //左上区之前没搜索就搜索                    {                        act_search(a, key, left, mid_col - 1, top, mid_row - 1, col);                    }                    if (a[mid_row*col + left] == key)                    {                        printf("a[%d][%d] = %d\n", top, mid_col, a[top*col + mid_col]);                    }                }                else   //a[mid_row][left]<key  排除左上区，搜索包括所在行的左下区                {                    act_search(a, key, left, mid_col - 1, mid_row, bot, col);                }            }            else    //(a[mid_row][mid_col] < key)   中心值小于key，排除左上区            {                if (a[mid_row*col + right] <= key)    //右边中心值不大于key，排除右上区，检索右下区                {                    act_search(a, key, mid_col + 1, right, mid_row + 1, bot, col);                    if (a[mid_row*col + bot] == key)                    {                        printf("a[%d][%d] = %d/n", mid_col, right, a[mid_col*col + right]);                    }                }                else   //a[mid_row][bot] > key 右上区可能有可以，检索                {                    act_search(a, key, mid_col + 1, right, top, mid_row, col);                }                if (a[bot*col + mid_col] <= key)  // 底行中心值不大于key排除左下区                {                    if (a[mid_row*col + right] > key)   //右下区之前未检索就检索                    {                        act_search(a, key, mid_col + 1, right, mid_row + 1, bot, col);                    }                    if (a[bot*col + mid_col] == key)                    {                        printf("a[%d][%d] = %d\n", bot, mid_col, a[bot*col + mid_col]);                    }                }                else  //a[bot][mid_col] > key  左下下区可能存在key                {                    act_search(a, key, left, mid_col, mid_row + 1, bot, col);                }            }        }        else  //a[mid_row][mid_col] == key  中心值为key，排除左上右下区，搜索不包括所在行列的右上左下区        {            printf("a[%d][%d] = %d\n", mid_row, mid_col, a[mid_row*col + mid_col]);            act_search(a, key, left, mid_col - 1, mid_row + 1, bot, col);            act_search(a, key, mid_col + 1, right, top, mid_row - 1, col);        }    }}void search(int *a, int key, int rol, int cow){    int left = 0;           //左边界列下标    int right = cow - 1;    //右边界列下标     int top = 0;            //上边界行下标    int bot = rol - 1;      //下边界行下标    act_search(a, key, left, right, top, bot, cow);   //注意传值列数cow}int main(){    int a[][3] = { { 1, 3, 4 }, { 2, 4, 5 }, { 4, 5, 6 } };   //初始化数组    int rol = sizeof(a) / sizeof(a[0]);                       //求行数    int cow = sizeof(a[0]) / sizeof(a[0][0]);                 //求列数    int key = 4;                                              //设要找的数    search(a[0], key, rol, cow);                             // 传参，搜索    return 0;}

运行结果相同

测试下稍复杂的数组