java - .equals()方法和==相等判断有什么区别？Leetcode 155中的问题

问题描述

问 题

本题是实现stack相关内容。
最早的时候写的代码如下，加粗部分的判断用==连接,

if(minstack.peek()==(stack.peek())

但是结果不正确，报告如下

Input:
["MinStack","push","push","push","push","pop","getMin","pop","getMin","pop","getMin"]
[[],[512],[-1024],[-1024],[512],[],[],[],[],[],[]]
Output:
["constructor","null","null","null","null","null","-1024","null","-1024","null","-1024"]
Expected:
["constructor","null","null","null","null","null","-1024","null","-1024","null","512"]

之后将加粗部分改为:

if(minstack.peek().equals(stack.peek()))

使用equals()连接，答案就是正确的了。

新人入码坑，求老师们指点一二这是为何？

另:这个结果报告是怎么看的，有点看不懂，是和二叉树有关的写法？求各位前辈指教！

谢谢！

附源码如下:

public class MinStack {
    Stack<Integer> minstack = new Stack<Integer>();
    Stack<Integer> stack = new Stack<Integer>();
   
    public MinStack() {     
    }
    
    public void push(int x) {
        if(minstack.empty() || x <= minstack.peek())
            minstack.push(x);
        stack.push(x);
    }
    public void pop() {
        **if(minstack.peek()==(stack.peek()))**
            minstack.pop();
        stack.pop();
    }
    public int top() {
        return stack.peek();   
    }
    public int getMin(){
        return minstack.peek();  
    }
}

解决方案

既然你要用stack来实现，还要用==来判断，那就应该push的时候做控制，让两个stack加入的对象是同一个。

public class MinStack {
    Stack<Integer> minstack = new Stack<Integer>();
    Stack<Integer> stack = new Stack<Integer>();
   
    public MinStack() {     
    }
    
    public void push(int x) {
        Integer val = new Integer(x);
        if(minstack.empty() || val <= minstack.peek())
            minstack.push(val);
        stack.push(val);
    }
    public void pop() {
        if(minstack.peek()==(stack.peek()))
            minstack.pop();
        stack.pop();
    }
    public int top() {
        return stack.peek();   
    }
    public int getMin(){
        return minstack.peek();  
    }
}

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