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:
1 class MyQueue { 2 public: 3 stack<int> stack1; 4 stack<int> stack2; 5 6 MyQueue() { 7 } 8 9 void push(int element) { 10 stack1.push(element); 11 } 12 13 void adjust() { 14 if (stack2.empty()) { 15 while (!stack1.empty()) { 16 stack2.push(stack1.top()); 17 stack1.pop(); 18 } 19 } 20 } 21 22 int pop() { 23 adjust(); 24 int temp = stack2.top(); 25 stack2.pop(); 26 return temp; 27 } 28 29 int top() { 30 adjust(); 31 return stack2.top(); 32 } 33 };