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 minstack = new Stack(); Stack stack = new Stack(); 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(); } }

public class MinStack { Stack minstack = new Stack(); Stack stack = new Stack(); 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(); } }

String a = "abc"; String b = "abc";

String a = "abc"; String b = new String("abc");