qq_44331767的博客

public static void createCut(Node headA,Node headB){ headA.next=headB.next.next; } //输入两个链表，找到链表相交的节点 public static Node getInterSectionNode(Node headA,Node headB){ Node ps=headA;//默认指向长的单链表 Node pl=headB;//默认指向指向短的单

public static Node mergeTwoLists(Node headA,Node headB){ Node newHead=new Node(-1); Node tmp=newHead; while(headA!=null&&headB!=null){ if(headA.data<headB.data){ tmp.next=headA; .

//找到环的入口 slow=this.head; slow=slow.next; fast=fast.next;public ListNode detectCycle() { ListNode fast = this.head; ListNode slow = this.head; while (fast != null && fast.next != null) { fast = fast.next.next; slow

//链表是否有环 //走三步，走一步 很慢才能相遇 ，或者永远不相遇 public boolean hasCycle() { Node fast = this.head; Node slow = this.head; while (fast != null && fast.next != null) { //可以快速相遇 fast = fast.next.next;

//链表是否是回文 找到链表的中间节点 反转单链表 head从后往后走 slow从后往前走 public boolean chkPalindrome() { if (this.head == null){ return false; } if (this.head.next == null){ return true; } Node fast=this.head;

//删除链表中重复的节点 public Node delete(){ Node newHead=new Node(-1); Node tmp=newHead; Node cur=this.head; while(cur!=null) { if (cur.next != null && cur.data == cur.next.data) {

/以x为基准，将链表分成两部分，原来的数据顺序不变 public Node partition(int x){ if(head==null){ return null; } Node cur=this.head; Node bs=null; Node be=null; Node as=null; Node ae=null; while(cur!=n

//链表中倒数第K各节点 k的合法性 public Node node(int k){ if(head==null){ return null; } if(k<=0){ System.out.println("K位置不合法"); return null; } Node fast=this.he

//返回链表的中间结点，遍历一遍 public Node midNode(){ Node fast=this.head; Node slow=this.head; while(fast!=null&&fast.next!=null){ fast=fast.next.next; slow=slow.next; } return slow; }

//反转一个单链表 只需遍历一遍 public Node reverse () { Node cur=this.head;//待反转的节点 Node prev=null; Node newHead=null; while(cur!=null){ Node curNext=cur.next; if(curNext==null