
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 int
-
- typedef 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);
- }