As the title described, you should only use two stacks to implement a queue's actions.
The queue should support push(element)
, pop()
and top()
where pop is pop the first(a.k.a front) element in the queue.
Both pop and top methods should return the value of first element.
Example
push(1)
pop() // return 1
push(2)
push(3)
top() // return 2
pop() // return 2
Challenge
implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.
解法一:
class MyQueue {
public:
stack<int> stack1;
stack<int> stack2; MyQueue() {
} void push(int element) {
stack1.push(element);
} void adjust() {
if (stack2.empty()) {
while (!stack1.empty()) {
stack2.push(stack1.top());
stack1.pop();
}
}
} int pop() {
adjust();
int temp = stack2.top();
stack2.pop();
return temp;
} int top() {
adjust();
return stack2.top();
}
};