剑指 Offer 09. 用两个栈实现队列
本问题对应的 leetcode 原文链接：剑指 Offer 09. 用两个栈实现队列
问题描述
用两个栈实现一个队列。队列的声明如下，请实现它的两个函数 appendTail 和 deleteHead ，分别完成在队列尾部插入整数和在队列头部删除整数的功能。(若队列中没有元素，deleteHead 操作返回 -1 )
示例 1：
输入：
["CQueue","appendTail","deleteHead","deleteHead","deleteHead"]
[[],[3],[],[],[]]
输出：[null,null,3,-1,-1]
示例 2:
输入：
["CQueue","deleteHead","appendTail","appendTail","deleteHead","deleteHead"]
[[],[],[5],[2],[],[]]
输出：[null,-1,null,null,5,2]
限制：
1 <= values <= 10000- 最多会对
appendTail、deleteHead进行10000次调用
解题思路
视频讲解直达： 本题视频讲解
代码实现
class CQueue {
Stack< Integer > stack1;
Stack< Integer > stack2;
public CQueue() {
stack1 = new Stack<>();
stack2 = new Stack<>();
}
public void appendTail(int value) {
stack1.push(value);
}
public int deleteHead() {
if(!stack2.isEmpty()){
return stack2.pop();
}
if(!stack1.isEmpty()){
while(!stack1.isEmpty()){
stack2.push(stack1.pop());
}
return stack2.pop();
}
return -1;
}
}
Python
class CQueue(object):
def __init__(self):
self.stack1 = []
self.stack2 = []
def appendTail(self, value):
"""
:type value: int
:rtype: None
"""
self.stack1.append(value)
def deleteHead(self):
"""
:rtype: int
"""
if self.stack2:
return self.stack2.pop()
if self.stack1:
while self.stack1:
self.stack2.append(self.stack1.pop())
return self.stack2.pop()
return -1
# Your CQueue object will be instantiated and called as such:
# obj = CQueue()
# obj.appendTail(value)
# param_2 = obj.deleteHead()
C++
class CQueue {
public:
CQueue() {
}
void appendTail(int value) {
stack1.push(value);
}
int deleteHead() {
if (!stack2.empty()) {
int res = stack2.top();
stack2.pop();
return res;
}
if (!stack1.empty()) {
while (!stack1.empty()) {
stack2.push(stack1.top());
stack1.pop();
}
int res = stack2.top();
stack2.pop();
return res;
}
return -1;
}
private:
stack<int> stack1;
stack<int> stack2;
};
/**
* Your CQueue object will be instantiated and called as such:
* CQueue* obj = new CQueue();
* obj->appendTail(value);
* int param_2 = obj->deleteHead();
*/
Go
type CQueue struct {
stack1 []int
stack2 []int
}
func Constructor() CQueue {
return CQueue{}
}
func (this *CQueue) AppendTail(value int) {
this.stack1 = append(this.stack1, value)
}
func (this *CQueue) DeleteHead() int {
if len(this.stack2) > 0 {
res := this.stack2[len(this.stack2)-1]
this.stack2 = this.stack2[:len(this.stack2)-1]
return res
}
if len(this.stack1) > 0 {
for len(this.stack1) > 0 {
this.stack2 = append(this.stack2, this.stack1[len(this.stack1)-1])
this.stack1 = this.stack1[:len(this.stack1)-1]
}
res := this.stack2[len(this.stack2)-1]
this.stack2 = this.stack2[:len(this.stack2)-1]
return res
}
return -1
}
/**
* Your CQueue object will be instantiated and called as such:
* obj := Constructor();
* obj.AppendTail(value);
* param_2 := obj.DeleteHead();
*/
JS
/**
* @constructor
*/
var CQueue = function() {
this.stack1 = [];
this.stack2 = [];
};
/**
* @param {number} value
* @return {void}
*/
CQueue.prototype.appendTail = function(value) {
this.stack1.push(value);
};
/**
* @return {number}
*/
CQueue.prototype.deleteHead = function() {
if(!this.stack2.length){
while(this.stack1.length){
this.stack2.push(this.stack1.pop());
}
}
if(!this.stack2.length){
return -1;
}else{
return this.stack2.pop();
}
};
/**
* Your CQueue object will be instantiated and called as such:
* var obj = new CQueue()
* obj.appendTail(value)
* var param_2 = obj.deleteHead()
*/
评论(1)
class CQueue:
时间复杂度： O(1)，push 为 O(1)，pop 为均摊 O(1)。对于每个元素，至多入栈和出栈各两次（Stack_In和Stack_Out至多各进出栈一次），故均摊复杂度为 O(1)。
空间复杂度：O(n)。其中 n 是操作总数。对于有 n 次 push 操作的情况，队列中会有 n 个元素，故空间复杂度为 O(n)。