设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈绝对值？相对值

小结：

1、

常数时间内检索到最小元素

2、存储

存储绝对值？相对值

存储差异

3、

java-ide-debug

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

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。

push(x) -- 将元素 x 推入栈中。
pop() -- 删除栈顶的元素。
top() -- 获取栈顶元素。
getMin() -- 检索栈中的最小元素。

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();

巴特西

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈绝对值？相对值

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巴特西

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈 绝对值？相对值

最新文章

热门文章

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈绝对值？相对值