0641. 设计循环双端队列【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 循环数组

1. 📝 题目描述

• leetcode

设计实现双端队列。

实现 MyCircularDeque 类:

• MyCircularDeque(int k) ：构造函数,双端队列最大为 k。
• boolean insertFront()：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。
• boolean insertLast() ：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。
• boolean deleteFront() ：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。
• boolean deleteLast() ：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。
• int getFront() ：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 getRear() ：获得双端队列的最后一个元素。 如果双端队列为空，返回 -1。
• boolean isEmpty() ：若双端队列为空，则返回 true，否则返回 false。
• boolean isFull() ：若双端队列满了，则返回 true，否则返回 false。

示例 1：

输入
["MyCircularDeque", "insertLast", "insertLast", "insertFront", "insertFront", "getRear", "isFull", "deleteLast", "insertFront", "getFront"]
[[3], [1], [2], [3], [4], [], [], [], [4], []]
输出
[null, true, true, true, false, 2, true, true, true, 4]

解释
MyCircularDeque circularDeque = new MycircularDeque(3); // 设置容量大小为3
circularDeque.insertLast(1);			        // 返回 true
circularDeque.insertLast(2);			        // 返回 true
circularDeque.insertFront(3);			        // 返回 true
circularDeque.insertFront(4);			        // 已经满了，返回 false
circularDeque.getRear();  				// 返回 2
circularDeque.isFull();				        // 返回 true
circularDeque.deleteLast();			        // 返回 true
circularDeque.insertFront(4);			        // 返回 true
circularDeque.getFront();				// 返回 4

提示：

• 1 <= k <= 1000
• 0 <= value <= 1000
• insertFront, insertLast, deleteFront, deleteLast, getFront, getRear, isEmpty, isFull 调用次数不大于 2000 次

2. 🎯 s.1 - 循环数组

c

typedef struct {
    int* arr;
    int cap;
    int front;
    int rear;
} MyCircularDeque;

MyCircularDeque* myCircularDequeCreate(int k) {
    MyCircularDeque* obj = (MyCircularDeque*)malloc(sizeof(MyCircularDeque));
    obj->cap = k + 1;
    obj->arr = (int*)malloc(sizeof(int) * obj->cap);
    obj->front = 0;
    obj->rear = 0;
    return obj;
}

bool myCircularDequeInsertFront(MyCircularDeque* obj, int value) {
    if ((obj->rear + 1) % obj->cap == obj->front) return false;
    obj->front = (obj->front - 1 + obj->cap) % obj->cap;
    obj->arr[obj->front] = value;
    return true;
}

bool myCircularDequeInsertLast(MyCircularDeque* obj, int value) {
    if ((obj->rear + 1) % obj->cap == obj->front) return false;
    obj->arr[obj->rear] = value;
    obj->rear = (obj->rear + 1) % obj->cap;
    return true;
}

bool myCircularDequeDeleteFront(MyCircularDeque* obj) {
    if (obj->front == obj->rear) return false;
    obj->front = (obj->front + 1) % obj->cap;
    return true;
}

bool myCircularDequeDeleteLast(MyCircularDeque* obj) {
    if (obj->front == obj->rear) return false;
    obj->rear = (obj->rear - 1 + obj->cap) % obj->cap;
    return true;
}

int myCircularDequeGetFront(MyCircularDeque* obj) {
    if (obj->front == obj->rear) return -1;
    return obj->arr[obj->front];
}

int myCircularDequeGetRear(MyCircularDeque* obj) {
    if (obj->front == obj->rear) return -1;
    return obj->arr[(obj->rear - 1 + obj->cap) % obj->cap];
}

bool myCircularDequeIsEmpty(MyCircularDeque* obj) {
    return obj->front == obj->rear;
}

bool myCircularDequeIsFull(MyCircularDeque* obj) {
    return (obj->rear + 1) % obj->cap == obj->front;
}

void myCircularDequeFree(MyCircularDeque* obj) {
    free(obj->arr);
    free(obj);
}

js

/**
 * @param {number} k
 */
var MyCircularDeque = function (k) {
  this.cap = k + 1
  this.arr = new Array(this.cap)
  this.front = 0
  this.rear = 0
}

MyCircularDeque.prototype.insertFront = function (value) {
  if (this.isFull()) return false
  this.front = (this.front - 1 + this.cap) % this.cap
  this.arr[this.front] = value
  return true
}

MyCircularDeque.prototype.insertLast = function (value) {
  if (this.isFull()) return false
  this.arr[this.rear] = value
  this.rear = (this.rear + 1) % this.cap
  return true
}

MyCircularDeque.prototype.deleteFront = function () {
  if (this.isEmpty()) return false
  this.front = (this.front + 1) % this.cap
  return true
}

MyCircularDeque.prototype.deleteLast = function () {
  if (this.isEmpty()) return false
  this.rear = (this.rear - 1 + this.cap) % this.cap
  return true
}

MyCircularDeque.prototype.getFront = function () {
  if (this.isEmpty()) return -1
  return this.arr[this.front]
}

MyCircularDeque.prototype.getRear = function () {
  if (this.isEmpty()) return -1
  return this.arr[(this.rear - 1 + this.cap) % this.cap]
}

MyCircularDeque.prototype.isEmpty = function () {
  return this.front === this.rear
}

MyCircularDeque.prototype.isFull = function () {
  return (this.rear + 1) % this.cap === this.front
}

py

class MyCircularDeque:
    def __init__(self, k: int):
        self.cap = k + 1
        self.arr = [0] * self.cap
        self.front = 0
        self.rear = 0

    def insertFront(self, value: int) -> bool:
        if self.isFull():
            return False
        self.front = (self.front - 1) % self.cap
        self.arr[self.front] = value
        return True

    def insertLast(self, value: int) -> bool:
        if self.isFull():
            return False
        self.arr[self.rear] = value
        self.rear = (self.rear + 1) % self.cap
        return True

    def deleteFront(self) -> bool:
        if self.isEmpty():
            return False
        self.front = (self.front + 1) % self.cap
        return True

    def deleteLast(self) -> bool:
        if self.isEmpty():
            return False
        self.rear = (self.rear - 1) % self.cap
        return True

    def getFront(self) -> int:
        return -1 if self.isEmpty() else self.arr[self.front]

    def getRear(self) -> int:
        return -1 if self.isEmpty() else self.arr[(self.rear - 1) % self.cap]

    def isEmpty(self) -> bool:
        return self.front == self.rear

    def isFull(self) -> bool:
        return (self.rear + 1) % self.cap == self.front

• 时间复杂度：$O(1)$，所有操作均为常数时间
• 空间复杂度：$O(k)$

算法思路：

• 用大小为 k + 1 的循环数组，维护 front 和 rear 两个指针
• 空判断：front == rear；满判断：(rear + 1) % cap == front
• 头部插入向前移动 front，尾部插入向后移动 rear