【LeetCode】232. Implement Queue using Stacks
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
, andis 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).
题解:
用一个辅助栈逆序,不过每push一个元素就要反转整个栈,非常耗时,不可取
Solution 1
class MyQueue { public: /** Initialize your data structure here. */ MyQueue() { } /** Push element x to the back of queue. */ void push(int x) { stack<int> tmp; while(!s.empty()){ tmp.push(s.top()); s.pop(); } s.push(x); while(!tmp.empty()){ s.push(tmp.top()); tmp.pop(); } } /** Removes the element from in front of queue and returns that element. */ int pop() { int tmp = s.top(); s.pop(); return tmp; } /** Get the front element. */ int peek() { return s.top(); } /** Returns whether the queue is empty. */ bool empty() { return s.empty(); } stack<int> s; };
设两个栈,一个只能push,一个只能pop。有些情况需要注意,只有s2栈空了,才能把s1栈元素push进s2,而且一定要全部push进s2。另外在返回首元素时,有时候是刚开始push进s1,还没有pop操作(也就是说s1的元素还没有push进s2,此时s2为空,但是s1不为空),那么就要和pop操作一样要将栈1元素全部push进栈2。
Solution 2
class MyQueue { public: /** Initialize your data structure here. */ MyQueue() { } /** Push element x to the back of queue. */ void push(int x) { s1.push(x); } /** Removes the element from in front of queue and returns that element. */ int pop() { if(s2.empty() && s1.empty()){ return INT_MIN; } else if(s2.empty()){ while(!s1.empty()){ int tmp = s1.top(); s1.pop(); s2.push(tmp); } } int tmp = s2.top(); s2.pop(); return tmp; } /** Get the front element. */ int peek() { if(!s2.empty()) return s2.top(); while(!s1.empty()){ int tmp = s1.top(); s1.pop(); s2.push(tmp); } return s2.top(); } /** Returns whether the queue is empty. */ bool empty() { return s1.empty() && s2.empty(); } stack<int> s1, s2; };