LeetCode -- Min Stack
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode -- Min Stack相关的知识,希望对你有一定的参考价值。
Question:
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.
Analysis:
设计一个栈,它能够支持push, pop, top操作,并且能够在常数时间内检索到最小的元素。
solution1: 利用Java内置的Stack类,并且设置一个格外的栈,当每次有元素进栈时都判断是否是额外栈的最小元素。这样当求最小元素时直接peek额外栈的栈顶元素即可。
solution2: 不适用Java内置的Stack类,每个栈节点维护一个最小值(是当前节点至栈底的最小值),其余操作与上面相同,只不过push,pop等基本操作要自己简单的实现一下。
Answer:
Solution1:
class MinStack { private Stack<Integer> stack = new Stack<Integer>(); private Stack<Integer> minStack = new Stack<Integer>(); public void push(int x) { if(minStack.isEmpty() || x <= minStack.peek()) //if x is the minimum value minStack.push(x); stack.push(x); } public void pop() { if(stack.peek().equals(minStack.peek())) //if the pop value is the minimum minStack.pop(); stack.pop(); } public int top() { return stack.peek(); } public int getMin() { return minStack.peek(); } }
Solution2:
class MinStack { StackNode top = null; public void push(int x) { if(top == null) { top = new StackNode(x); top.min = x; top.next = null; } else { StackNode node = new StackNode(x); if(x < top.min) node.min = x; else node.min = top.min; node.next = top; top = node; } } public void pop() { if(top != null) top = top.next; } public int top() { if(top != null) return top.val; else return 0; } public int getMin() { if(top != null) return top.min; else return 0; } } class StackNode { int val; int min; //store the min utile this node StackNode next; public StackNode(int x) { val = x; } }
以上是关于LeetCode -- Min Stack的主要内容,如果未能解决你的问题,请参考以下文章