本文介绍了C中的数组旋转的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试旋转如下所示的数组:
I am trying to rotate an array that looks like this:
a a a a
a b a a
b b b a
a a a a
我应该将其旋转5次,共90度。
I am supposed to rotate it 5 times for 90 degrees. It is supposed to be done in C.
我非常感谢每一个帮助,因为我只是一个初学者,而且对此一无所知。
I appreciate every help because I am just a beginner and am stuck on this.
提前感谢。
#include <stdio.h>
int main()
{
char array_1[4][4] = { {'-','-','-','-'},
{'-','o','-','-'},
{'o','o','o','-'},
{'-','-','-','-'}};
char array_2[4][4] = { {'-','-','-','-'},
{'-','o','o','-'},
{'o','o','-','-'},
{'-','-','-','-'}};
char array_3[4][4] = { {'-','-','-','-'},
{'-','o','-','-'},
{'-','o','-','-'},
{'-','o','o','-'}};
char array_4[4][4] = { {'-','-','o','-'},
{'-','-','o','-'},
{'-','-','o','-'},
{'-','-','o','-'}};
int counter = 0;
int counter_1 = 0;
for(counter = 0; counter < 4; counter++)
{
for(counter_1 = 0; counter_1 < 4; counter_1++)
{
printf("%c ",array_1[counter][counter_1]);
}
printf(" ");
for(counter_1 = 0; counter_1 < 4; counter_1++)
{
printf("%c ",array_2[counter][counter_1]);
}
printf(" ");
for(counter_1 = 0; counter_1 < 4; counter_1++)
{
printf("%c ",array_3[counter][counter_1]);
}
printf(" ");
for(counter_1 = 0; counter_1 < 4; counter_1++)
{
printf("%c ",array_4[counter][counter_1]);
}
printf(" ");
printf("\n");
}
printf("\n");
for(counter= 0; counter < 4; counter++)
{
for(counter_1 = 3; counter_1 >= 0; counter_1--)
{
printf("%c ",array_1[counter_1][counter]);
}
printf(" ");
for(counter_1 = 3; counter_1 >= 0; counter_1--)
{
printf("%c ",array_2[counter_1][counter]);
}
printf(" ");
for(counter_1 = 3; counter_1 >= 0; counter_1--)
{
printf("%c ",array_3[counter_1][counter]);
}
printf(" ");
for(counter_1 = 3; counter_1 >= 0; counter_1--)
{
printf("%c ",array_4[counter_1][counter]);
}
printf(" ");
printf("\n");
}
printf("\n");
推荐答案
像这样:
#include <stdio.h>
typedef struct point { int x, y; } Point;
void rotate(int n, char array[n][n]){
//rotate right 90 degrees
if(n == 1) return ;
int times = n / 2;
for(int i = 0; i < times; ++i){
Point base = { i, i };
for(int j = 0; j < n - 1; ++j){
Point transition[4] = { {j, n-1}, {n-1,n-1-j},{n-1-j,0},{0,j} };
char curr = array[base.x][base.y+j];//base + {0,j}
for(int k = 0; k < 4; ++k){
char temp = array[base.x + transition[k].x][base.y + transition[k].y];
array[base.x + transition[k].x][base.y + transition[k].y] = curr;
curr = temp;
}
}
n -= 2;
}
}
void display(int n, char array[n][n]){
for(int i = 0; i < n; ++i){
for(int j = 0; j < n; ++j){
if(j)
putchar(' ');
putchar(array[i][j]);
}
putchar('\n');
}
putchar('\n');
}
int main(void){
//demo
char array4[4][4] = {
{'1','2','3','4'},
{'5','6','7','8'},
{'9','A','B','C'},
{'D','E','F','0'}
};
display(4, array4);
int n = 4;
while(n--){
rotate(4, array4);
display(4, array4);
}
char array5[5][5] = {
{'A','B','C','D','E'},
{'F','G','H','I','J'},
{'K','L','M','N','O'},
{'P','Q','R','S','T'},
{'U','V','W','X','Y'}
};
display(5, array5);
n = 4;
while(n--){
rotate(5, array5);
display(5, array5);
}
}
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