Min Stack

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.

设计一个栈, 支持4种操作: push, pop, top和getMin, 关键是getMin是获取栈中的最小值.

示例:

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.

基本代码框架:

class MinStack {

    /** initialize your data structure here. */
    public MinStack() {

    }

    public void push(int x) {

    }

    public void pop() {

    }

    public int top() {

    }

    public int getMin() {

    }
}

在线测试网站: https://leetcode.com/problems/min-stack/description/

class MinStack {
    //存gap, 当前栈顶和之前的最小值的差值
    Stack<Long> stack = new Stack<>();
    long min = Integer.MIN_VALUE;
    public void push(int x) {
        if (stack.isEmpty()) {
            stack.push(0L);//初始值存0
            min = x;
        } else {
            stack.push(x - min);//存gap
            if (x < min) {
                min = x;
            }
        }
    }

    public void pop() {
        if (stack.isEmpty()) {
            return;
        }
        long peek = stack.peek();
        if (peek > 0) {
            stack.pop();
        } else {
            min -= peek;
            stack.pop();
        }
    }

    public int top() {
        if (stack.isEmpty()) {
            return -1;
        }
        if (stack.peek() > 0) {
            return (int)(min + stack.peek());
        } else {
            return (int)min;
        }
    }

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