Leetcode-155. Min Stack

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

博客链接：mcf171的博客

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

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

Example:

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

这个题目不难，但是自己做的时候有一个没考虑，就是堆里面存的是对象，对象比较不能用等于号。 Your runtime beats 23.66% of java submissions.

public class MinStack {

    Stack<Integer> data,min;

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

    public void push(int x) {
        data.push(x);
        if(min.empty() || x <= min.peek()) min.push(x);
    }

    public void pop() {
        int datanum = data.peek();
        int minnum = min.peek();
        if(datanum == minnum) min.pop();
        data.pop();

    }

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

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

使用equals方法也可以 Your runtime beats 30.04% of java submissions.

public class MinStack {

    Stack<Integer> data,min;

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

    public void push(int x) {
        data.push(x);
        if(min.empty() || x <= min.peek()) min.push(x);
    }

    public void pop() {
        if(data.peek().equals(min.peek())) min.pop();
        data.pop();

    }

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

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