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 解题图解：

1、 先用stack1存储push来的数据

2、每当要pop数据时，从stack2中取，如果 stack2为空，就先从stack1中“倒”数据到stack2。

 这就是用栈实现队列的基本操作

 这道题看起来比较容易，但是！如果你用C语言去去做，你还得自己去写一个栈，栈不知道写的可以看看之前的一篇文章——数据结构（C语言）——栈的两种实现方式

本题使用 动态数组构造栈

#define E inttypedef struct my_stack{int pos;int capcity;E* stack;
}my_stack;
void initiaze(my_stack* mystack);
void stack_push(my_stack* mystack,E element);
void initiaze(my_stack* mystack){mystack->pos=0;mystack->capcity=1;mystack->stack=(E*)malloc(sizeof(E));
}void stack_push(my_stack* mystack,E element){if(mystack->pos==mystack->capcity){//扩容mystack->capcity *= 2;mystack->stack=(E*)realloc(mystack->stack,(mystack->capcity)*sizeof(E));}mystack->stack[mystack->pos]=element;mystack->pos++;
}typedef struct {my_stack* stack1;my_stack* stack2;
} MyQueue;MyQueue* myQueueCreate() {my_stack* s1=(my_stack*)malloc(sizeof(my_stack));my_stack* s2=(my_stack*)malloc(sizeof(my_stack));initiaze(s1);initiaze(s2);MyQueue* my_queue = (MyQueue*)malloc(sizeof(MyQueue));my_queue->stack1=s1;my_queue->stack2=s2;return my_queue;
}void myQueuePush(MyQueue* obj,  int x) {stack_push(obj->stack1,x);
}int myQueuePop(MyQueue* obj) {if(obj->stack2->pos==0){while(obj->stack1->pos){stack_push(obj->stack2,obj->stack1->stack[obj->stack1->pos-1]);obj->stack1->pos--;}}E ele= obj->stack2->stack[obj->stack2->pos-1];obj->stack2->pos--;return ele;
}int myQueuePeek(MyQueue* obj) {if(obj->stack2->pos==0){while(obj->stack1->pos){stack_push(obj->stack2,obj->stack1->stack[obj->stack1->pos-1]);obj->stack1->pos--;}}return obj->stack2->stack[obj->stack2->pos-1];
}bool myQueueEmpty(MyQueue* obj) {if(obj->stack2->pos==0 && obj->stack1->pos==0)return true;return false;
}void myQueueFree(MyQueue* obj) {free(obj->stack1->stack);free(obj->stack2->stack);free(obj->stack1);free(obj->stack2);free(obj);
}