[剑指offer][JAVA]面试题第[30]题[包含min函数的栈][双栈辅助栈][单栈]

【问题描述】[中等]

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。示例:MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.min();   --> 返回 -2.

【解答思路】

1. 单栈

时间复杂度：O(N^2) 空间复杂度：O(1)

在这里插入图片描述

class MinStack {private LinkedList<Integer> data;private int min;/** initialize your data structure here. */public MinStack() {data = new LinkedList<Integer>();min = Integer.MAX_VALUE;}public void push(int x) {if(x <= min){data.add(min);min = x;}data.add(x);}public void pop() {if(data.pollLast() == min){//flush the minmin = data.pollLast();}}public int top() {return data.peekLast();}public int min() {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.min();*/

Stack

class MinStack {private Stack<Integer> stack;private int min;/** initialize your data structure here. */public MinStack() {stack = new Stack<>();min = Integer.MAX_VALUE;}public void push(int x) {if(x <= min ){//注意：这里要使用<=号stack.push(min);//在每一个min入栈之前将它前一个min入栈min = x;}stack.push(x);}public void pop() {if(stack.pop() == min){//如果取出来的是当前min，就将压在它低下的前一个min出栈min = stack.pop();}}public int top() {return stack.peek();}public int min() {return min;}
}

2. 双栈（辅助栈）

在这里插入图片描述

时间复杂度：O(1) 空间复杂度：O(N)

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

class MinStack {Stack<Integer> A, B;public MinStack() {A = new Stack<>();B = new Stack<>();}public void push(int x) {A.add(x);if(B.empty() || B.peek() >= x)B.add(x);}public void pop() {if(A.pop().equals(B.peek()))B.pop();}public int top() {return A.peek();}public int min() {return B.peek();}
}

【总结】

1.Stack 的常用方法：

push( num) 入栈
pop() 栈顶元素出栈
empty() 判定栈是否为空
peek() 获取栈顶元素
search(num) 判端元素num是否在栈中，如果在返回1，不在返回-1。    注意pop()和peek()的区别。pop()会弹出栈顶元素并返回栈顶的值，peek()只是获取栈顶的值，但是并不会把元素从栈顶弹出来