155. Min Stack

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.

链接： http://leetcode.com/problems/min-stack/

题解：

最小栈。这道题在电面Bloomberg的supply chain组的Senior SDE时遇到了原题，秒杀掉了。然后面试的老印很坏，follow up说你这个不是stack，说想要new stack<>()的时候自带min功能，跟我绕了半天，故意误导，好恶心...最后果然电面没过-_____-!! 后来看了CC150, 才知道这老印也许想要的是CC150的写法 - MinStack extends Stack<Integer>，然后用minStack和super进行互动，也是两个栈。

又聊远了，这道题目，保持stack的功能同时还要有getMin()。一般的方法是维护两个stack。也可以维护一个stack，但要创建一个Node class，空间复杂度并不能被节省。还有的做法是用一个stack，stack里存min到当前值x的距离，然后有些计算，比较巧妙。

维护两个stack， Time Complexity (pop，push，getMin，top) - O(1) ， Space Complexity - O(n)。

class MinStack {
    private Stack<Integer> stack = new Stack<>();
    private Stack<Integer> minStack = new Stack<>();
    
    public void push(int x) {
        if(minStack.isEmpty() || x <= minStack.peek())
            minStack.push(x);
        stack.push(x);
    }

    public void pop() {
        if(stack.peek().equals(minStack.peek()))
            minStack.pop();
        stack.pop();
    }

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

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

维护一个栈， Time Complexity (pop，push，getMin，top) - O(1) ， Space Complexity - O(n)。

class MinStack {
    private Stack<Node> stack;
    
    public MinStack() {
        this.stack = new Stack<>();    
    }
    
    public void push(int x) {
        int min = Math.min(x, getMin());
        this.stack.push(new Node(x, min));
    }

    public void pop() {
        this.stack.pop();
    }

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

    public int getMin() {
        if(this.stack.isEmpty())
            return Integer.MAX_VALUE;
        return this.stack.peek().min;
    }
    
    private class Node {
        public int min;
        public int val;
        
        public Node(int val, int min) {
            this.val = val;
            this.min = min;
        }
    }
}

二刷：

除了维护两个栈的方法以外，还有两种方法，一种是建立一个Node同时保存当前值和最小值。

另外一种是reeclapple的写法。

设定一个min，push的时候，除了第一个数外，只把每个数x和min的差值(x - min)存入栈。当 x< min的时候更新min = x
pop的时候pop出这个差值，假如这个值小于0，我们更新min = min - pop，否则min不变
计算peek的时候，先peek出栈顶元素，假如这个元素top大于0，那么返回 top + min，否则返回min。这里理解比较tricky.
计算getMin的时候直接返回min

Java:

最简单的维护两个stack的。

class MinStack {
    private Stack<Integer> minStack = new Stack<>();
    private Stack<Integer> stack = new Stack<>();
    
    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty() || x <= minStack.peek()) {
            minStack.push(x);
        }
    }

    public void pop() {
        int x = stack.pop();
        if (x == minStack.peek()) {
            minStack.pop();
        }
    }

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

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

三刷:

要注意假如遇到的话，还要加上各种边界条件判断以及throw EmptyStackException

Java:

class MinStack {
    private Stack<Integer> stack = new Stack<>();
    private Stack<Integer> minStack = new Stack<>();
    
    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty() || x <= minStack.peek()) minStack.push(x);
    }

    public void pop() {
        int x = stack.pop();
        if (x == minStack.peek()) minStack.pop();
    }

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

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

使用一个Stack:

每次min要被更新的时候，先存入min，再存入新元素。相当于保持一个递减的min序列。假如elements递减，则栈内元素类似 {min 1, elem1, min2, elem2, min3, elem3}等等，

这里在insert的时候，假如x <= min，则我们先把之前被更新过的min放进stack里，把min设为当前的x，之后把x放入栈
在pop的时候，我们pop出一个元素，然后跟min比较，假如等于min，那说明下一个元素是旧的min，我们更新min为 stack.pop()，也把这个旧的min pop出来
top 直接返回 stack.peek();
getMin直接返回min

class MinStack {
    private Stack<Integer> stack = new Stack<>();
    int min = Integer.MAX_VALUE;
    
    public void push(int x) {
        if (x <= min) {
            stack.push(min);
            min = x;
        }
        stack.push(x);
    }

    public void pop() {
        int peek = stack.pop();
        if (peek == min) min = stack.pop();
    }

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

    public int getMin() {
        return min;
    }
}

Update:

许多题目现在必须追求最优解了 -____-!!

public class MinStack {
    private Stack<Integer> stack = new Stack<>();
    private int min = Integer.MAX_VALUE

    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<>();
    }
    
    public void push(int x) {
        if (x <= min) {
            stack.push(min);
            min = x;
        }
        stack.push(x);
    }
    
    public void pop() {
        int top = stack.pop();
        if (top == min) min = stack.pop();
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return min;
    }
}

/**
 * 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();
 */

Reference：

http://www.geeksforgeeks.org/design-and-implement-special-stack-data-structure/

https://leetcode.com/discuss/21071/java-accepted-solution-using-one-stack

https://leetcode.com/discuss/15679/share-my-java-solution-with-only-one-stack

https://leetcode.com/discuss/19389/java-solution-accepted

https://leetcode.com/discuss/23927/smart-accepted-java-solution-linked-list

https://leetcode.com/discuss/16029/shortest-and-fastest-1-stack-and-2-stack-solutions

posted @ 2015-05-09 12:33 YRB 阅读(633) 评论(0) 编辑收藏举报

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155. Min Stack

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