题目描述

• 用空间换时间的经典例子，唯一要解决的地方就是 getMin() 如何实现常数检索
• （貌似也是面试常考题）
  

思路 & 代码

• 既然想快，那就要付出空间的代价
• 用什么空间可以实现？数组好找，但是不好扩展；链表好扩，但是不好找；同样是顺序结构，不可能用队列，那就用自己吧！来一个辅助栈
• 辅助栈：记录主栈每个元素为栈顶时，对应的最大值。

class MinStack {private Stack<Integer> stack;private Stack<Integer> helper;// 思路，维护一个最小栈helper，对应当前子栈的最小值/** initialize your data structure here. */public MinStack() {stack = new Stack<>();helper = new Stack<>();}public void push(int val) {stack.push(val);if(helper.empty()){helper.push(val);}else{helper.push(val > helper.peek() ? helper.peek() : val);}}public void pop() {stack.pop();helper.pop();}public int top() {return stack.peek();}public int getMin() {return helper.peek();}
}/*** Your MinStack object will be instantiated and called as such:* MinStack obj = new MinStack();* obj.push(val);* obj.pop();* int param_3 = obj.top();* int param_4 = obj.getMin();*/

更新版

• 换成 ArrayDeque，效率更高了

class MinStack {Deque<Integer> stack;Deque<Integer> minStack;/** initialize your data structure here. */public MinStack() {stack = new ArrayDeque<>();minStack = new ArrayDeque<>();}public void push(int val) {stack.push(val);int min = minStack.isEmpty() ? val : minStack.peek();val = val < min ? val : min;minStack.push(val);}public void pop() {stack.pop();minStack.pop();}public int top() {return stack.peek();}public int getMin() {return minStack.peek();}
}