[英]traversing singly linked list in Java
I have a simple Java question. 我有一个简单的Java问题。 As shown in the below code:
如下面的代码所示:
public static ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
ListNode slow = start, fast = start;
slow.next = head;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}
Start, fast and slow address the same object. 开始,快速和慢速寻址同一对象。 I don't get why "slow = slow.next;"
我不明白为什么“ slow = slow.next;” will not change the start object, but "slow.next = slow.next.next;"
不会更改起始对象,但是“ slow.next = slow.next.next;” will change the start object.
将更改起始对象。
slow
is a local variable, so changing its value to refer to a new instance of ListNode
doesn't affect the original list. slow
是一个局部变量,因此更改其值以引用ListNode
的新实例不会影响原始列表。
However, if slow
refers to a ListNode
that belongs to your list, changing slow.next
to refer to a new instance changes the state of your list. 但是,如果
slow
是指ListNode
属于你的列表,改变slow.next
指一个新的实例改变你的列表的状态。
It may be clearer if you use a setter to modify the next node : 如果使用setter修改下一个节点,则可能会更清楚:
slow.next = slow.next.next;
would be equivalent to : 相当于:
slow.setNext(slow.next.next);
So if slow
refers to a ListNode
that belongs to your list, changing its state is changing the state of your list. 因此,如果
slow
引用了属于您列表的ListNode
,则更改其状态即更改了列表的状态。
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