题目:数据逆置 代码部分分析: ar[0] -> temp[9] ar[1] -> temp[8] ar[2] -> temp[7] ar[3] -> temp[6] ar[4] -> temp[5] ar[5] -> temp[4] ar[6] -> temp[3] ar[7] -> temp[2] ar[8] -> temp[1] ar[9] -> temp[0]
ar[i] -> temp[10-1-i] |
//方法1:
#include<stdio.h>
void main(void){
int ar[10] = {0,1,2,3,4,5,6,7,8,9};
int temp[10] = {0};
int i;
printf("初始ar数组的值为:");
for(i = 0; i < 10; i++){
printf("%d ", ar[i]);
temp[10-1-i] = ar[i];
}
printf("\ntemp数组的值为:");
for(i = 0; i < 10; i++){
printf("%d ", temp[i]);
ar[i] = temp[i];
}
printf("\n逆置后ar数组的值为:");
for(i = 0; i < 10; i++){
printf("%d ", ar[i]);
}
printf("\n");
}
//方法2:按中轴对称
#include<stdio.h>
void main(void){
int ar[10] = {0,1,2,3,4,5,6,7,8,9};
int temp;
int i;
printf("初始ar数组的值为:");
for(i = 0; i < 10; i++){
printf("%d ", ar[i]);
}
printf("\n");
for(i = 0; i < 10/2; i++){
temp = ar[i];
ar[i] = ar[10-1-i];
ar[10-1-i] = temp;
}
printf("\n逆置后ar数组的值为:");
for(i = 0; i < 10; i++){
printf("%d ", ar[i]);
}
printf("\n");
}