Example 64: C language implementation finds the saddle point in a two-dimensional array, that is, the element at that position is the largest in the row and the smallest in the column. There may also be no saddle points.
Analysis: When readers look at the question, they must first understand what a saddle point is, first find the element with the largest value in a row, and then check whether it is the smallest value in the column, if it is, it is a saddle point, and output the saddle point; if If no, then find the largest number in the next row... If the largest number in each row is not a saddle point, this array has no saddle points.
Source code demo:
#include<stdio.h>//头文件
#define N 3 //宏定义
#define M 4 //宏定义
int main()//主函数
{
int i,j,k,a[N][M],max,maxj,flag;//定义整型变量和二维数组
printf("请输入数组：\n");//提示语句
for(i=0;i<N;i++)
{
for(j=0;j<M;j++)
{
scanf("%d",&a[i][j]);//往数组里存数
}
}
for(i=0;i<N;i++)
{
max=a[i][0]; //开始时假设a[i][0]最大
maxj=0; //将列号0赋给maxj保存
for(j=0;j<M;j++) //找出第i行中的最大数
{
if(a[i][j]>max)
{
max=a[i][j]; //将本行最大的数放在max中
maxj=j; //将最大数所在的列号存放在maxj中
}
}
flag=1; //先假设是鞍点，以flag为1代表
for(k=0;k<N;k++)
{
if(max>a[k][maxj])
{
//将最大的数和其同列元素相比
flag=0; //如果max不是同列最小，表示不是鞍点
continue;
}
}
if(flag)
{
printf("a[%d][%d]=%d\n",i,maxj,max); //输出鞍点的值和所在行列号
break;
}
}
if(!flag)
{
printf("鞍点不存在!\n");//提示语句
}
return 0;//主函数返回值为0
}
The compilation and running results are as follows:
请输入数组：
1 2 3 4
5 6 7 8
9 10 11 12
a[0][3]=4
--------------------------------
Process exited after 10.52 seconds with return value 0
请按任意键继续. . .
Above, if you see it and think it is helpful to you, please give Xiaolin a thumbs up and share it with the people around him, so that Xiaolin will also have the motivation to update, thank you fathers and villagers~
C language to find the saddle point in a two-dimensional array
More cases can go to the public account: C language entry to proficient