链接:https://leetcode.cn/problems/design-circular-deque/solution/by-xun-ge-v-lht6/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。


解题思路
题意就是让我们实现一个循环双链表。
具体实现
根据题意直接模拟即可
注意几个细节点,题目需要围绕头节点实现循环双链表,那么我们可以额外申请一个头节点,用来保存循环双链表的头位置,并保存一些基本信息,比如链表长度。
同时可以和单链表一样,申请一个虚拟头结点,避免插入和删除操作时对头结点的分析

- typedef struct QueueData{
- int val;
- struct QueueData * next;
- struct QueueData * upper;
- }QueueData;
-
- typedef struct MyCircularDeque{
- int k;
- struct QueueData * top;
- } MyCircularDeque;
-
- //构造函数,双端队列最大为 k 。
- MyCircularDeque* myCircularDequeCreate(int k) {
- printf("create\t");
- MyCircularDeque * obj = malloc(sizeof(MyCircularDeque));
- obj->k = k;
- QueueData * data = malloc(sizeof(QueueData));
- data->val = -1;
- data->next = data;
- data->upper = data;
- obj->top = data;
-
- return obj;
- }
- //将一个元素添加到双端队列头部。 如果操作成功返回 true ,否则返回 false 。
- bool myCircularDequeInsertFront(MyCircularDeque* obj, int value) {
- printf("insertFront\t");
- if(!obj->k)
- return false;
- obj->k--;
- QueueData * data = malloc(sizeof(QueueData));
- data->val = value;
- data->next = obj->top->next;
- data->upper = obj->top;
- obj->top->next->upper = data;
- obj->top->next = data;
- return true;
- }
- //将一个元素添加到双端队列尾部。如果操作成功返回 true ,否则返回 false 。
- bool myCircularDequeInsertLast(MyCircularDeque* obj, int value) {
- printf("insertlast\t");
- if(!obj->k)
- return false;
- obj->k--;
- QueueData * data = malloc(sizeof(QueueData));
- data->val = value;
- data->next = obj->top;
- data->upper = obj->top->upper;
- obj->top->upper->next = data;
- obj->top->upper = data;
- return true;
- }
- //从双端队列头部删除一个元素。 如果操作成功返回 true ,否则返回 false 。
- bool myCircularDequeDeleteFront(MyCircularDeque* obj) {
- printf("dalafrint\t");
- if(obj->top->next != obj->top)
- {
- obj->k++;
- QueueData * n = obj->top->next;
- obj->top->next = obj->top->next->next;
- obj->top->next->upper = obj->top;
- free(n);
- return true;
- }
- return false;
- }
- //从双端队列尾部删除一个元素。如果操作成功返回 true ,否则返回 false 。
- bool myCircularDequeDeleteLast(MyCircularDeque* obj) {
- printf("daletelast\t");
- if(obj->top->upper != obj->top)
- {
- obj->k++;
- QueueData * n = obj->top->upper;
- obj->top->upper = obj->top->upper->upper;
- obj->top->upper->next = obj->top;
- free(n);
- return true;
- }
- return false;
- }
- //从双端队列头部获得一个元素。如果双端队列为空,返回 -1 。
- int myCircularDequeGetFront(MyCircularDeque* obj) {
- printf("getfront\t");
- if(obj->top->next != obj->top)
- {
- return obj->top->next->val;
- }
- return -1;
- }
- //获得双端队列的最后一个元素。 如果双端队列为空,返回 -1 。
- int myCircularDequeGetRear(MyCircularDeque* obj) {
- printf("getrear\t");
- if(obj->top->upper != obj->top)
- {
- return obj->top->upper->val;
- }
- return -1;
- }
- //若双端队列为空,则返回 true ,否则返回 false 。
- bool myCircularDequeIsEmpty(MyCircularDeque* obj) {
- printf("isempty\t");
- if(obj->top->next != obj->top)
- {
- return false;
- }
- return true;
- }
- //若双端队列满了,则返回 true ,否则返回 false 。
- bool myCircularDequeIsFull(MyCircularDeque* obj) {
- printf("isfull\t");
- if(obj->k)
- {
- return false;
- }
- return true;
- }
-
- void myCircularDequeFree(MyCircularDeque* obj) {
- obj->top->upper->next = NULL;
- for (QueueData *curr = obj->top;curr;) {
- printf("%d ",curr->val);
- QueueData *node = curr;
- curr = curr->next;
- free(node);
- }
- free(obj);
- return;
- }
-
- /**
- * Your MyCircularDeque struct will be instantiated and called as such:
- * MyCircularDeque* obj = myCircularDequeCreate(k);
- * bool param_1 = myCircularDequeInsertFront(obj, value);
-
- * bool param_2 = myCircularDequeInsertLast(obj, value);
-
- * bool param_3 = myCircularDequeDeleteFront(obj);
-
- * bool param_4 = myCircularDequeDeleteLast(obj);
-
- * int param_5 = myCircularDequeGetFront(obj);
-
- * int param_6 = myCircularDequeGetRear(obj);
-
- * bool param_7 = myCircularDequeIsEmpty(obj);
-
- * bool param_8 = myCircularDequeIsFull(obj);
-
- * myCircularDequeFree(obj);
- */
-
- 作者:xun-ge-v
- 链接:https://leetcode.cn/problems/design-circular-deque/solution/by-xun-ge-v-lht6/
- 来源:力扣(LeetCode)
- 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。