Question

C语言二维数组元素怎么“顺时针旋转”，如图

Answer 1

#include"stdio.h"

int  main()

{int i,j,n,t,a[20][20];

int c;

scanf("%d",&n);

for(i=0;i<n;i++)

{for(j=0;j<n;j++)

{a[i][j]=n*i+j+1;

printf("%3d",a[i][j]);

}

printf("\n");

}

printf("After:\n");

for(i=0;i<n-1;i++)

for(j=i;j<n-1-i;j++)

{t=a[i][j];

a[i][j]=a[n-1-j][i];

a[n-1-j][i]=a[n-1-i][n-1-j];

a[n-1-i][n-1-j]=a[j][n-1-i];

a[j][n-1-i]=t;

}

for(i1=0;i1<n;i1++)

{for(j1=0;j1<n;j1++)

printf("%3d",a[i1][j1]);

printf("\n");

}

return 0;

}

Answer 2

//关键是理解表达式b[i][j] = a[2 - j][i];
//我用的c++，只有输入和输出和c不同，逻辑一样
#include<iostream>
using namespace std;
int main() {
int a[3][3] = { {1,2,3} ,{4,5,6}, {7,8,9} };
cout << "转化之前的数组为：" << endl;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
cout << a[i][j] << " ";
}
cout << endl;
}
int b[3][3];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
//逆时针b[i][j] = a[j][2 - i];
b[i][j] = a[2 - j][i];
}

}
cout << "转化之后的数组为：" << endl;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
cout << b[i][j] << " ";
}
cout << endl;
}

}

追问

十分感谢大佬