LeetCode 716. 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.