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

标签：操作   leetcode   ==   could   turn   str   push   operation   ati   

小结：

1、

常数时间内检索到最小元素

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.

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;
        }
    }
}

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

标签：操作   leetcode   ==   could   turn   str   push   operation   ati   

原文地址：https://www.cnblogs.com/yuanjiangw/p/10801228.html