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155. 最小栈

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题目描述

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.

解法

Python3

class MinStack:

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


    def push(self, x: int) -> None:
        self.s.append(x)
        element = x if not self.helper or x < self.helper[-1] else self.helper[-1]
        self.helper.append(element)

    def pop(self) -> None:
        self.s.pop()
        self.helper.pop()

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

    def getMin(self) -> int:
        return self.helper[-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.getMin()

Java

class MinStack {

    private Deque<Integer> s;
    private Deque<Integer> helper;

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

    public void push(int x) {
        s.push(x);
        int element = helper.isEmpty() || x < helper.peek() ? x : helper.peek();
        helper.push(element);
    }

    public void pop() {
        s.pop();
        helper.pop();
    }

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

    public int getMin() {
        return helper.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.getMin();
 */

...