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面试题 03.02. 栈的最小值

English Version

题目描述

请设计一个栈，除了常规栈支持的pop与push函数以外，还支持min函数，该函数返回栈元素中的最小值。执行push、pop和min操作的时间复杂度必须为O(1)。

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

利用辅助栈存放栈的最小元素。

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 len(self._s2) == 0 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 getMin(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.getMin()

Java

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

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

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

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

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

    public int getMin() {
        return s2.empty() ? -1 : 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.getMin();
 */

...