剑指 Offer 30. 包含min函数的栈

问题描述（原题链接）

  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)。

代码：

超时

class MinStack {

    /** initialize your data structure here. */
    Stack<Integer> stack1;
    Stack<Integer> stack2;
    Stack<Integer> stack3;
    public MinStack() {
        stack1 = new Stack<Integer>();//存取数据
        stack2 = new Stack<Integer>();//最小元素
        stack3 = new Stack<Integer>();//临时
    }
    
    public void push(int x) {
        stack1.push(x);
        if(stack2.isEmpty()){
            stack2.push(x);
        }else{
            if(x<=stack2.peek())
            stack2.push(x);
            else{
                while(!stack2.isEmpty() && stack2.peek()<x){
                    stack3.push(stack2.pop());
                }
                stack2.push(x);
                while(!stack3.isEmpty()){
                    stack2.push(stack3.pop());
                }
            }
        }
    }
    
    public void pop() {
        int temp = stack1.peek();
        stack1.pop();
        while(!stack2.isEmpty()&&stack2.peek()!=temp){
            stack3.push(stack2.pop());
        }
        stack2.pop();
        while(!stack3.isEmpty()){
            stack2.push(stack3.pop());
        }
    }
    
    public int top() {
        return stack1.peek();
    }
    
    public int min() {
        return stack2.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();
 */

改进

class MinStack {

    /** initialize your data structure here. */
    Stack<Integer> stack1;
    Stack<Integer> stack2;
    public MinStack() {
        stack1 = new Stack<Integer>();//存取数据
        stack2 = new Stack<Integer>();//最小元素
    }
    
    public void push(int x) {
        stack1.push(x);
        if(stack2.isEmpty()){
            stack2.push(x);
        }else{
            if(x<=stack2.peek())
            stack2.push(x);
            else{
                stack2.push(stack2.peek());
            }
        }
    }
    
    public void pop() {
        stack1.pop();
        stack2.pop();
    }
    
    public int top() {
        return stack1.peek();
    }
    
    public int min() {
        return stack2.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();
 */