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