面试题 30. 包含 min 函数的栈

题目描述

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。

示例:

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

提示：

• 各函数的调用总次数不超过 20000 次

解法

Python3

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.s1 = []
        self.s2 = []

    def push(self, x: int) -> None:
        self.s1.append(x)
        self.s2.append(x if not self.s2 or self.s2[-1] >= x else self.s2[-1])

    def pop(self) -> None:
        self.s1.pop()
        self.s2.pop()

    def top(self) -> int:
        return self.s1[-1]

    def min(self) -> int:
        return self.s2[-1]


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()

Java

class MinStack {
    private Deque<Integer> s1;
    private Deque<Integer> s2;

    /** initialize your data structure here. */
    public MinStack() {
        s1 = new ArrayDeque<>();
        s2 = new ArrayDeque<>();
    }

    public void push(int x) {
        s1.push(x);
        s2.push(s2.isEmpty() || s2.peek() >= x ? x : s2.peek());
    }

    public void pop() {
        s1.pop();
        s2.pop();
    }

    public int top() {
        return s1.peek();
    }

    public int min() {
        return s2.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.min();
 */

JavaScript

/**
 * initialize your data structure here.
 */
var MinStack = function () {
  this.stack = [];
  this.minStack = [];
};

/**
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function (x) {
  this.stack.unshift(x);
  if (!this.minStack.length || this.minStack[0] >= x) {
    this.minStack.unshift(x);
  }
};

/**
 * @return {void}
 */
MinStack.prototype.pop = function () {
  if (this.stack.shift() === this.minStack[0]) {
    this.minStack.shift();
  }
};

/**
 * @return {number}
 */
MinStack.prototype.top = function () {
  return this.stack[0];
};

/**
 * @return {number}
 */
MinStack.prototype.min = function () {
  return this.minStack[0];
};

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

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