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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.s = []
        self.mins = [float('inf')]

    def push(self, val: int) -> None:
        self.s.append(val)
        self.mins.append(min(self.mins[-1], val))

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

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

    def getMin(self) -> int:
        return self.mins[-1]


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

Java

class MinStack {
    private Deque<Integer> s;
    private Deque<Integer> mins;

    /** initialize your data structure here. */
    public MinStack() {
        s = new ArrayDeque<>();
        mins = new ArrayDeque<>();
        mins.push(Integer.MAX_VALUE);
    }
    
    public void push(int val) {
        s.push(val);
        mins.push(Math.min(mins.peek(), val));
    }
    
    public void pop() {
        s.pop();
        mins.pop();
    }
    
    public int top() {
        return s.peek();
    }
    
    public int getMin() {
        return mins.peek();
    }
}

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

...