【LeetCode】Min Stack 解题报告

【题目】

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.

来自：https://oj.leetcode.com/discuss/15659/simple-java-solution-using-two-build-in-stacks?state=edit-15691&show=15659#q15659

class MinStack {
    // stack: store the stack numbers
    private Stack<Integer> stack = new Stack<Integer>();
    // minStack: store the current min values
    private Stack<Integer> minStack = new Stack<Integer>();

    public void push(int x) {
        // store current min value into minStack
        if (minStack.isEmpty() || x <= minStack.peek())
            minStack.push(x);
        stack.push(x);
    }

    public void pop() {
        // use equals to compare the value of two object, if equal, pop both of them
        if (stack.peek().equals(minStack.peek()))
            minStack.pop();
        stack.pop();
    }

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

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

这道题的关键之处就在于 minStack 的设计，push() pop() top() 这些操作Java内置的Stack都有，不必多说。

我最初想着再弄两个数组，分别记录每一个元素的前一个比它大的和后一个比它小的，想复杂了。

第一次看上面的代码，还认为它有问题，为啥仅仅在 x<minStack.peek() 时压栈？假设，push(5), push(1), push(3) 这样minStack里不就仅仅有5和1，这样pop()出1后， getMin() 不就得到5而不是3吗？事实上这样想是错的，由于要想pop()出1之前，3就已经被pop()出了。. 

minStack 记录的永远是当前全部元素中最小的，不管 minStack.peek() 在stack 中所处的位置。

【不用内置Stack的实现】

来自：https://oj.leetcode.com/discuss/15651/my-java-solution-without-build-in-stack

class MinStack {
    Node top = null;

    public void push(int x) {
        if (top == null) {
            top = new Node(x);
            top.min = x;
        } else {
            Node temp = new Node(x);
            temp.next = top;
            top = temp;
            top.min = Math.min(top.next.min, x);
        }
    }

    public void pop() {
        top = top.next;
        return;
    }

    public int top() {
        return top == null ? 0 : top.val;
    }

    public int getMin() {
        return top == null ? 0 : top.min;
    }
}

class Node {
    int val;
    int min;
    Node next;

    public Node(int val) {
        this.val = val;
    }
}