作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/design-circular-deque/description/

题目描述:

Design your implementation of the circular double-ended queue (deque).
Your implementation should support following operations:

  • MyCircularDeque(k): Constructor, set the size of the deque to be k.
  • insertFront(): Adds an item at the front of Deque. Return true if the operation is successful.
  • insertLast(): Adds an item at the rear of Deque. Return true if the operation is successful.
  • deleteFront(): Deletes an item from the front of Deque. Return true if the operation is successful.
  • deleteLast(): Deletes an item from the rear of Deque. Return true if the operation is successful.
  • getFront(): Gets the front item from the Deque. If the deque is empty, return -1.
  • getRear(): Gets the last item from Deque. If the deque is empty, return -1.
  • isEmpty(): Checks whether Deque is empty or not.
  • isFull(): Checks whether Deque is full or not.

Example:

MyCircularDeque circularDeque = new MycircularDeque(3); // set the size to be 3
circularDeque.insertLast(1); // return true
circularDeque.insertLast(2); // return true
circularDeque.insertFront(3); // return true
circularDeque.insertFront(4); // return false, the queue is full
circularDeque.getRear(); // return 32
circularDeque.isFull(); // return true
circularDeque.deleteLast(); // return true
circularDeque.insertFront(4); // return true
circularDeque.getFront(); // return 4

Note:

  • All values will be in the range of [1, 1000].
  • The number of operations will be in the range of [1, 1000].
  • Please do not use the built-in Deque library.

题目大意

实现一个双向环形链表。

解题方法

使用列表

622. Design Circular Queue很类似了,622题要求的是单向的环形链表,这个题要求是双向的。

做法同样是采用和622题目类似的“作弊”做法,用一个list实现类似的环形列表。因为是双向链表,所以可以从头尾插入元素,对应了list的insert和append方法。

特别注意的是,插入元素,无论是在头尾插入,要移动的指针都是rear。

我这个题的做法比622更加成熟一点,622是通过头部指针来实现的,这个题里面是直接操作的头部元素导致后面的元素跟着变化,头部指针没有用到。。

这么一说可以把头部指针删除了23333.

在这么一说发现rear一直指向的是list的结尾,所以也可以删除了23333.

没删除头部指针的代码如下:

class MyCircularDeque(object):

    def __init__(self, k):
"""
Initialize your data structure here. Set the size of the deque to be k.
:type k: int
"""
self.queue = []
self.size = k
self.front = 0
self.rear = 0 def insertFront(self, value):
"""
Adds an item at the front of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.insert(0, value)
self.rear += 1
return True
else:
return False def insertLast(self, value):
"""
Adds an item at the rear of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.append(value)
self.rear += 1
return True
else:
return False def deleteFront(self):
"""
Deletes an item from the front of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop(0)
self.rear -= 1
return True
else:
return False def deleteLast(self):
"""
Deletes an item from the rear of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop()
self.rear -= 1
return True
else:
return False def getFront(self):
"""
Get the front item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[self.front] def getRear(self):
"""
Get the last item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[self.rear -1] def isEmpty(self):
"""
Checks whether the circular deque is empty or not.
:rtype: bool
"""
return self.front == self.rear def isFull(self):
"""
Checks whether the circular deque is full or not.
:rtype: bool
"""
return self.rear - self.front == self.size # Your MyCircularDeque object will be instantiated and called as such:
# obj = MyCircularDeque(k)
# param_1 = obj.insertFront(value)
# param_2 = obj.insertLast(value)
# param_3 = obj.deleteFront()
# param_4 = obj.deleteLast()
# param_5 = obj.getFront()
# param_6 = obj.getRear()
# param_7 = obj.isEmpty()
# param_8 = obj.isFull()

其实可以删除头指针,因为头指针一直指向0.

删除头部指针的代码如下:

class MyCircularDeque(object):

    def __init__(self, k):
"""
Initialize your data structure here. Set the size of the deque to be k.
:type k: int
"""
self.queue = []
self.size = k
self.rear = 0 def insertFront(self, value):
"""
Adds an item at the front of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.insert(0, value)
self.rear += 1
return True
else:
return False def insertLast(self, value):
"""
Adds an item at the rear of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.append(value)
self.rear += 1
return True
else:
return False def deleteFront(self):
"""
Deletes an item from the front of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop(0)
self.rear -= 1
return True
else:
return False def deleteLast(self):
"""
Deletes an item from the rear of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop()
self.rear -= 1
return True
else:
return False def getFront(self):
"""
Get the front item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[0] def getRear(self):
"""
Get the last item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[self.rear -1] def isEmpty(self):
"""
Checks whether the circular deque is empty or not.
:rtype: bool
"""
return 0 == self.rear def isFull(self):
"""
Checks whether the circular deque is full or not.
:rtype: bool
"""
return self.rear == self.size # Your MyCircularDeque object will be instantiated and called as such:
# obj = MyCircularDeque(k)
# param_1 = obj.insertFront(value)
# param_2 = obj.insertLast(value)
# param_3 = obj.deleteFront()
# param_4 = obj.deleteLast()
# param_5 = obj.getFront()
# param_6 = obj.getRear()
# param_7 = obj.isEmpty()
# param_8 = obj.isFull()

rear 始终等于 len(self.queue),所以完全可以删除。

删除front和rear指针的代码如下:

class MyCircularDeque(object):

    def __init__(self, k):
"""
Initialize your data structure here. Set the size of the deque to be k.
:type k: int
"""
self.queue = []
self.size = k def insertFront(self, value):
"""
Adds an item at the front of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.insert(0, value)
return True
else:
return False def insertLast(self, value):
"""
Adds an item at the rear of Deque. Return true if the operation is successful.
:type value: int
:rtype: bool
"""
if not self.isFull():
self.queue.append(value)
return True
else:
return False def deleteFront(self):
"""
Deletes an item from the front of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop(0)
return True
else:
return False def deleteLast(self):
"""
Deletes an item from the rear of Deque. Return true if the operation is successful.
:rtype: bool
"""
if not self.isEmpty():
self.queue.pop()
return True
else:
return False def getFront(self):
"""
Get the front item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[0] def getRear(self):
"""
Get the last item from the deque.
:rtype: int
"""
if self.isEmpty():
return -1
else:
return self.queue[-1] def isEmpty(self):
"""
Checks whether the circular deque is empty or not.
:rtype: bool
"""
return 0 == len(self.queue) def isFull(self):
"""
Checks whether the circular deque is full or not.
:rtype: bool
"""
return self.size == len(self.queue) # Your MyCircularDeque object will be instantiated and called as such:
# obj = MyCircularDeque(k)
# param_1 = obj.insertFront(value)
# param_2 = obj.insertLast(value)
# param_3 = obj.deleteFront()
# param_4 = obj.deleteLast()
# param_5 = obj.getFront()
# param_6 = obj.getRear()
# param_7 = obj.isEmpty()
# param_8 = obj.isFull()

双向链表

上面的解法直接使用了内置数据结构list,在头部插入的效率比较慢,时间复杂度是O(N)!所以我使用了双向链表模拟deque.

维护双向链表的方式和STL的list一样。对啊,可以直接用List。。方法就不仔细讲了。

struct Node {
Node* prev;
Node* next;
int val;
Node(int v) : val(v) {};
};
struct List {
Node* dummy;
int size;
List() : size(0) {
dummy = new Node(0);
dummy->next = dummy;
dummy->prev = dummy;
};
};
class MyCircularDeque {
public:
/** Initialize your data structure here. Set the size of the deque to be k. */
MyCircularDeque(int k) {
maxSize = k;
} /** Adds an item at the front of Deque. Return true if the operation is successful. */
bool insertFront(int value) {
if (list.size == maxSize)
return false;
Node* cur = new Node(value);
Node* prev = list.dummy->prev;
list.dummy->prev = cur;
cur->prev = prev;
prev->next = cur;
cur->next = list.dummy;
++list.size;
return true;
} /** Adds an item at the rear of Deque. Return true if the operation is successful. */
bool insertLast(int value) {
if (list.size == maxSize)
return false;
Node* cur = new Node(value);
Node* next = list.dummy->next;
list.dummy->next = cur;
cur->next = next;
next->prev = cur;
cur->prev = list.dummy;
++list.size;
return true;
} /** Deletes an item from the front of Deque. Return true if the operation is successful. */
bool deleteFront() {
if (list.size == 0) return false;
Node* prev = list.dummy->prev;
Node* pprev = prev->prev;
list.dummy->prev = pprev;
pprev->next = list.dummy;
--list.size;
return true;
} /** Deletes an item from the rear of Deque. Return true if the operation is successful. */
bool deleteLast() {
if (list.size == 0) return false;
Node* next = list.dummy->next;
Node* nnext = next->next;
list.dummy->next = nnext;
nnext->prev = list.dummy;
--list.size;
return true;
} /** Get the front item from the deque. */
int getFront() {
if (list.size == 0) return -1;
return list.dummy->prev->val;
} /** Get the last item from the deque. */
int getRear() {
if (list.size == 0) return -1;
return list.dummy->next->val;
} /** Checks whether the circular deque is empty or not. */
bool isEmpty() {
return list.size == 0;
} /** Checks whether the circular deque is full or not. */
bool isFull() {
return list.size == maxSize;
}
private:
List list;
int maxSize;
}; /**
* Your MyCircularDeque object will be instantiated and called as such:
* MyCircularDeque obj = new MyCircularDeque(k);
* bool param_1 = obj.insertFront(value);
* bool param_2 = obj.insertLast(value);
* bool param_3 = obj.deleteFront();
* bool param_4 = obj.deleteLast();
* int param_5 = obj.getFront();
* int param_6 = obj.getRear();
* bool param_7 = obj.isEmpty();
* bool param_8 = obj.isFull();
*/

日期

2018 年 7 月 13 日 —— 早起困一上午,中午必须好好休息才行啊
2019 年 2 月 26 日 —— 二月就要完了

【LeetCode】641. Design Circular Deque 解题报告(Python & C++)的更多相关文章

  1. [LeetCode] 641.Design Circular Deque 设计环形双向队列

    Design your implementation of the circular double-ended queue (deque). Your implementation should su ...

  2. LeetCode 641. Design Circular Deque

    原题链接在这里:https://leetcode.com/problems/design-circular-deque/ 题目: Design your implementation of the c ...

  3. 【LeetCode】622. Design Circular Queue 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 用直的代替弯的 数组循环利用 日期 题目地址:htt ...

  4. LC 641. Design Circular Deque

    Design your implementation of the circular double-ended queue (deque). Your implementation should su ...

  5. 【LeetCode】62. Unique Paths 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/unique-pa ...

  6. [LeetCode] 622.Design Circular Queue 设计环形队列

    Design your implementation of the circular queue. The circular queue is a linear data structure in w ...

  7. 【LeetCode】649. Dota2 Senate 解题报告(Python)

    [LeetCode]649. Dota2 Senate 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地 ...

  8. 【LeetCode】886. Possible Bipartition 解题报告(Python)

    [LeetCode]886. Possible Bipartition 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu ...

  9. 【LeetCode】870. Advantage Shuffle 解题报告(Python)

    [LeetCode]870. Advantage Shuffle 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn ...

随机推荐

  1. Excel—分组然后取每组中对应时间列值最大的或者最小的

    1.MAX(IF(A:A=D2,B:B)) 输入函数公式后,按Ctrl+Shift+Enter键使函数公式成为数组函数公式. Ctrl+Shift+Enter: 按住Ctrl键不放,继续按Shift键 ...

  2. excel--CLEAN()函数,解决为什么看着相同的字符串但是len()长度不同

    CLEAN()函数能够有效解决去除字符串中隐藏的字符(这些字符是TRIM()去除不掉的)

  3. 设置administrator账号密码

    设置administrator账号密码: 打开:附件->运行 输入:lusrmgr.msc 在里面的用户里修改administrator密码

  4. GraphScope 集群部署

    GraphScope 集群部署 1 k8s集群搭建 大致步骤如下: 安装docker.在ubuntu上,可以简单的通过命令sudo apt install docker.io来安装. 安装kubele ...

  5. HTTP请求 Java API

    1.导入依赖 <dependency> <groupId>commons-httpclient</groupId> <artifactId>common ...

  6. iOS调用系统电话、浏览器、地图、邮件等

    - (IBAction)openMaps { //打开地图 NSString*addressText = @"beijing"; //@"1Infinite Loop, ...

  7. 使用 ACE 库框架在 UNIX 中开发高性能并发应用

    使用 ACE 库框架在 UNIX 中开发高性能并发应用来源:developerWorks 中国 作者:Arpan Sen ACE 开放源码工具包可以帮助开发人员创建健壮的可移植多线程应用程序.本文讨论 ...

  8. 【Java 8】Predicate详解

    一.java.util.function.Predicate 这里类是java自带主要广泛用在支持lambda表达式的API中. 1.接口源码 @FunctionalInterface public ...

  9. 第一章-Flink介绍-《Fink原理、实战与性能优化》读书笔记

    Flink介绍-<Fink原理.实战与性能优化>读书笔记 1.1 Apache Flink是什么? 在当代数据量激增的时代,各种业务场景都有大量的业务数据产生,对于这些不断产生的数据应该如 ...

  10. 修复Apache Log4j任意代码执行漏洞安全风险通告

    2021年12月10日 0x01漏洞背景 Apache Log4j 是 Apache 的一个开源项目,Apache Log4j2是一个基于Java的日志记录工具.该工具重写了Log4j框架,并且引入了 ...