【一天一道LeetCode】#232. Implement Queue using Stacks
一天一道LeetCode
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(一)题目
Implement the following operations of a queue using stacks.
push(x) – Push element x to the back of queue.
pop() – Removes the element from in front of queue.
peek() – Get the front element.
empty() – Return whether the queue is empty.
Notes:
+ You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid.
+ Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
+ You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue)
(二)解题
题目大意:用两个栈实现queue,要求实现push,pop,peek和empty
解题思路:栈是先进后出,队列是先进先出。利用两个栈stack1和stack2
push:和栈一样,直接push到一个栈stack1中
pop:有一个辅助栈就可以将stack1中的数依次push到stack2中,这样stack2的top就是最先进来的数,直接pop即可
peek:去队首的数,如果stack2不为空的话,直接返回stack2.pop即可,如果stack2为空,则先把stack1的数push过来,再去top元素
empty:如果stack1和stack2均为空,就直接返回true,否则返回false
具体实现如下:
class Queue {
public:
// Push element x to the back of queue.
void push(int x) {
st1.push(x);
}
// Removes the element from in front of queue.
void pop(void) {
if(empty()) return;
if(st2.empty())
{
while(!st1.empty()){//为空的话要把stack1的元素push进来
st2.push(st1.top());
st1.pop();
}
}
st2.pop();
}
// Get the front element.
int peek(void) {
if(st2.empty()){//为空的话要把stack1的元素push进来
while(!st1.empty()){
st2.push(st1.top());
st1.pop();
}
}
return st2.top();
}
// Return whether the queue is empty.
bool empty(void) {
if(st1.empty()&&st2.empty()) return true;
else return false;
}
private:
stack<int> st1;
stack<int> st2;
};
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