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[leetcode]包含min函数的栈 min stack

admin 11月 11, 2020 0

题目描述

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.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

代码如下：

public class MinStack {

    Stack<Integer> stack1;
    Stack<Integer> stack2;
    
    
    public MinStack() {
        stack1 = new Stack<>();
        //存储stack的最小值
        stack2 = new Stack<>();
    }
    
    public void push(int x) {
        stack1.push(x);
        if(stack2.isEmpty() ){
            stack2.push(x);
        }else{
            if(stack2.peek() >= x) stack2.push(x);
        }
    }
    
    public void pop() {
        if(stack2.peek().equals(stack1.peek())) stack2.pop();
        stack1.pop();
        
    }
    
    public int top() {
        return stack1.peek();
    }
    
    public int getMin() {
        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.getMin();
 */