LeetCode Max Stack

原题链接在这里：https://leetcode.com/problems/max-stack/description/

题目：

Design a max stack that supports push, pop, top, peekMax and popMax.

1. push(x) -- Push element x onto stack.
2. pop() -- Remove the element on top of the stack and return it.
3. top() -- Get the element on the top.
4. peekMax() -- Retrieve the maximum element in the stack.
5. popMax() -- Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.

Example 1:

MaxStack stack = new MaxStack();
stack.push(5);
stack.push(1);
stack.push(5);
stack.top(); -> 5
stack.popMax(); -> 5
stack.top(); -> 1
stack.peekMax(); -> 5
stack.pop(); -> 1
stack.top(); -> 5

Note:

1. -1e7 <= x <= 1e7
2. Number of operations won't exceed 10000.
3. The last four operations won't be called when stack is empty.

题解：

两个stack来解决. 不同的是popMax, 用temp stack 来保存找到最大值之前的, pop出max后再加回去, 同时更新maxStk.

Note: when popMax, get temp back, it needs to update maxStk as well.

Time Complexity: push, O(1). pop, O(1). top, O(1). peekMax, O(1). popMax, O(n). n是stack中数字的个数.

Space: O(n).

AC Java:

 class MaxStack {
     Stack<Integer> stk;
     Stack<Integer> maxStk;

     /** initialize your data structure here. */
     public MaxStack() {
         stk = new Stack<Integer>();
         maxStk = new Stack<Integer>();
     }

     public void push(int x) {
         stk.push(x);
         if(maxStk.isEmpty() || maxStk.peek()<=x){
             maxStk.push(x);
         }
     }

     public int pop() {
         int x = stk.pop();
         if(!maxStk.isEmpty() && x==maxStk.peek()){
             maxStk.pop();
         }

         return x;
     }

     public int top() {
         return stk.peek();
     }

     public int peekMax() {
         return maxStk.peek();
     }

     public int popMax() {
         Stack<Integer> tempStk = new Stack<Integer>();
         int x = maxStk.pop();
         while(!stk.isEmpty() && stk.peek()<x){
             tempStk.push(stk.pop());
         }

         stk.pop();
         while(!tempStk.isEmpty()){
             int top = tempStk.pop();
             push(top);
         }
         return x;
     }
 }

 /**
  * Your MaxStack object will be instantiated and called as such:
  * MaxStack obj = new MaxStack();
  * obj.push(x);
  * int param_2 = obj.pop();
  * int param_3 = obj.top();
  * int param_4 = obj.peekMax();
  * int param_5 = obj.popMax();
  */

类似Min Stack.