标签:strong line 数组 toc tle nbsp ret alt .com
??将a[0]与a[n-1]对换,再将a[1]与a[n-2]对换…直到将a[int(n-1)]与a[int((n-1)/2)-1]对换。
??如图所示:

//数组实现逆置void conver_arr(int c,int a[]){int low =0;int high = c -1;for(int i =0; i < c/2; i++){if(low < high){int temp = a[c - i -1]; a[c - i -1]= a[i]; a[i]= temp;}}}//指针实现逆置void conver_point(int c,int*a){int*start; start = a;int*end;end= a + c -1;while(start <end){int*temp =*start;*start =*end;*end= temp;*start++;*end--;}}//打印输出信息void arr_print(int*a){for(int i =0; i <10; i++){ printf("a[%d] = %d\n", i+1, a[i]);}}int main(void){int arr_value[10]={12,34,5,67,3,54,6,31,46,1};int len =sizeof(arr_value)/sizeof(arr_value[0]); conver_arr(len, arr_value); conver_point(len, arr_value); arr_print(arr_value);return0;}??运行结果如下图所示:

标签:strong line 数组 toc tle nbsp ret alt .com
原文地址:http://www.cnblogs.com/Bob-tong/p/6610833.html