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155. Min Stack - 最小栈

Tags - 题目标签

Description - 题目描述

EN:

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

Implement the MinStack class:

  • MinStack() initializes the stack object.
  • void push(int val) pushes the element val onto the stack.
  • void pop() removes the element on the top of the stack.
  • int top() gets the top element of the stack.
  • int getMin() retrieves the minimum element in the stack.

You must implement a solution with O(1) time complexity for each function.

 

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

 

Constraints:

  • -231 <= val <= 231 - 1
  • Methods pop, top and getMin operations will always be called on non-empty stacks.
  • At most 3 * 104 calls will be made to push, pop, top, and getMin.

ZH-CN:

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。

实现 MinStack 类:

  • MinStack() 初始化堆栈对象。
  • void push(int val) 将元素val推入堆栈。
  • void pop() 删除堆栈顶部的元素。
  • int top() 获取堆栈顶部的元素。
  • int getMin() 获取堆栈中的最小元素。

 

示例 1:

输入：
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

输出：
[null,null,null,null,-3,null,0,-2]

解释：
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.

 

提示：

  • -231 <= val <= 231 - 1
  • pop、top 和 getMin 操作总是在 非空栈 上调用
  • push, pop, top, and getMin最多被调用 3 * 104 次

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language Runtime Memory Submission Time
typescript 112 ms 51.3 MB 2023/03/04 22:03 
type StackItem = {
  value: number;
  minVal: number;
}

class MinStack {
  private _stack: StackItem[];
  private _min: number;

  constructor() {
    this._stack = [];
    this._min = +Infinity;
  }

  push(val: number): void {
    if (this._min > val) {
      this._min = val;
    }
    this._stack.push({
      value: val,
      minVal: this._min
    });
  }

  pop(): void {
    this._stack.pop();
    if (this._stack.length !== 0) {
      this._min = this._stack[this._stack.length - 1].minVal;
    } else {
      this._min = +Infinity;
    }
  }

  top(): number {
    return this._stack[this._stack.length - 1].value;
  }

  getMin(): number {
    return this._stack[this._stack.length - 1].minVal;
  }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * var obj = new MinStack()
 * obj.push(val)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */

My Notes - 我的笔记

stackItem 保存一个入栈时的最小值状态即可。

type StackItem = {
  value: number;
  minVal: number;
}

class MinStack {
  private _stack: StackItem[];
  private _min: number;

  constructor() {
    this._stack = [];
    this._min = +Infinity;
  }

  push(val: number): void {
    if (this._min > val) {
      this._min = val;
    }
    this._stack.push({
      value: val,
      minVal: this._min
    });
  }

  pop(): void {
    this._stack.pop();
    if (this._stack.length !== 0) {
      this._min = this._stack[this._stack.length - 1].minVal;
    } else {
      this._min = +Infinity;
    }
  }

  top(): number {
    return this._stack[this._stack.length - 1].value;
  }

  getMin(): number {
    return this._stack[this._stack.length - 1].minVal;
  }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * var obj = new MinStack()
 * obj.push(val)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */

注意： 栈清空时，需要重新把全局最小值设为+Infinity。