删除链表中的节点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
node.val=node.next.val;
node.next=node.next.next;
}
}
删除链表的倒数第N个节点
两次遍历算法
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy=new ListNode(0);
dummy.next=head;
int length=0;
ListNode p=head;
while (p!=null) {
length++;
p=p.next;
}
ListNode first=dummy;
length-=n;
while (length>0) {
length--;
first=first.next;
}
first.next=first.next.next;
return dummy.next;
}
}
一次遍历算法
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
// Advances first pointer so that the gap between first and second is n nodes apart
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
// Move first to the end, maintaining the gap
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}
反转链表
我的解法：利用Linklist比较好理解
class Solution {
public ListNode reverseList(ListNode head) {
LinkedList<Integer> list=new LinkedList<>();
for(ListNode x=head;x!=null;x=x.next) {
list.add(x.val);
}
for(ListNode x=head;x!=null;x=x.next) {
x.val=list.removeLast();
}
return head;
}
}