原标题:https://oj.leetcode.com/problems/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.
题目事实上不难。可是是leetCode今天刚刚新增的一个题目。索性抢个头彩把解题过程分享出来:
直接上AC代码:
public class MinStack { private int min = Integer.MAX_VALUE; private int minIndex = -1; private ArrayList<Integer> stack = new ArrayList<Integer>(); private int length = 0; private HashMap<Integer,Integer> secondMinMap = new HashMap<Integer, Integer>(); public void push(int x) { stack.add(x); if(x <= min) { //注意这里犯过错误,须要加=号,原因是当栈空的时候假设push一个Integer.MaxValue的话。须要将此时的minIndex变为0而不是继续为-1。//否则的话。这样的情况下。当push第二个数的时候,将出现secondMap.put(1,-1)的情况,进而当pop这个数的时候,会出现stack.get(-1)操作。
//换句话说。这个secondMap里的值仅仅有当key=0的时候。才干为-1,其它情况必须有能指向数组里位置的值 secondMinMap.put(length, minIndex); //这里存的 length位置相应的第二小的,仅仅考虑比length小的索引即可 //由于用到的场景是当前这个假设被pop了,说明它上面的全部都已经被pop了 //这个时候假设当前是min。那么接下来仅仅须要在它以下去找第二小的即可了 minIndex = length; min = x; } length ++; } public void pop() { if(length == 0) return; if(minIndex == length-1) { if(minIndex == 0) { minIndex = -1; min = Integer.MAX_VALUE; } else { minIndex = secondMinMap.get(length-1); secondMinMap.remove(length-1); min = stack.get(minIndex); } } stack.remove(length-1); length--; } public int top() { return stack.get(length-1); } public int getMin() { return min; } public static void main(String[] args) { MinStack stack = new MinStack(); stack.push(2147483646); stack.push(2147483646); stack.push(2147483647); System.out.println(stack.top()); stack.pop(); System.out.println(stack.getMin()); stack.pop(); System.out.println(stack.getMin()); stack.pop(); stack.push(2147483647); System.out.println(stack.top()); System.out.println(stack.getMin()); stack.push(-2147483648); System.out.println(stack.top()); System.out.println(stack.getMin()); stack.pop(); System.out.println(stack.getMin()); } }
版权声明:本文博主原创文章。博客,未经同意不得转载。