設計一個支持 push，pop，top 操作，並能在常數時間內檢索到最小元素的棧 絕對值？相對值

小結：

1、

常數時間內檢索到最小元素

 2、存儲

存儲絕對值？相對值

存儲差異

 3、

java-ide-debug

 

最小棧 - 力扣（LeetCode）
https://leetcode-cn.com/problems/min-stack/

設計一個支持 push，pop，top 操作，並能在常數時間內檢索到最小元素的棧。

• push(x) -- 將元素 x 推入棧中。
• pop() -- 刪除棧頂的元素。
• top() -- 獲取棧頂元素。
• getMin() -- 檢索棧中的最小元素。

示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.getMin();   --> 返回 -2.

 

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

 

package leetcode;

import java.util.Stack;

class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();

    public static void main(String[] args) {
        MinStack minStack = new MinStack();
        minStack.push(-2);
        minStack.push(0);
        minStack.push(-3);
        minStack.getMin();
        minStack.pop();
        minStack.top();
        minStack.getMin();
    }

    public void push(int x) {
        // only push the old minimum value when the current
        // minimum value changes after pushing the new value x
        if (x <= min) {
            stack.push(min);
            min = x;
        }
        stack.push(x);
    }

    public void pop() {
        // if pop operation could result in the changing of the current minimum value,
        // pop twice and change the current minimum value to the last minimum value.
        if (stack.pop() == min) min = stack.pop();
    }

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

    public int getMin() {
        return min;
    }
}

 

(3) Clean 6ms Java solution - LeetCode Discuss
https://leetcode.com/problems/min-stack/discuss/49010/Clean-6ms-Java-solution

不借助java stack

package leetcode;

class MinStack {
    private Node head;

    public void push(int x) {
        if (head == null)
            head = new Node(x, x);
        else
            head = new Node(x, Math.min(x, head.min), head);
    }

    public void pop() {
        head = head.next;
    }

    public int top() {
        return head.val;
    }

    public int getMin() {
        return head.min;
    }

    private class Node {
        int val;
        int min;
        Node next;

        private Node(int val, int min) {
            this(val, min, null);
        }

        private Node(int val, int min, Node next) {
            this.val = val;
            this.min = min;
            this.next = next;
        }
    }
}

 

 

(3) Share my Java solution with ONLY ONE stack - LeetCode Discuss
https://leetcode.com/problems/min-stack/discuss/49031/Share-my-Java-solution-with-ONLY-ONE-stack

The question is ask to construct One stack. So I am using one stack.

 

The idea is to store the gap between the min value and the current value;

 

The problem for my solution is the cast. I have no idea to avoid the cast. Since the possible gap between the current value and the min value could be Integer.MAX_VALUE-Integer.MIN_VALUE;

 

public class MinStack { long min; Stack<Long> stack; public MinStack(){ stack=new Stack<>(); } public void push(int x) { if (stack.isEmpty()){ stack.push(0L); min=x; }else{ stack.push(x-min);//Could be negative if min value needs to change if (x<min) min=x; } } public void pop() { if (stack.isEmpty()) return; long pop=stack.pop(); if (pop<0) min=min-pop;//If negative, increase the min value } public int top() { long top=stack.peek(); if (top>0){ return (int)(top+min); }else{ return (int)(min); } } public int getMin() { return (int)min; } }

 

 

 

It's brilliant to store only the difference!! But regarding pop, doesn't it return nothing? Isn't it defined as public void pop(){}
//long pop=stack.pop();