一篇文章搞定面试中的链表题目(java实现)

bianchenglangzi · · 1373 次点击 · 开始浏览    置顶
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![image.png](https://upload-images.jianshu.io/upload_images/17176526-d97300ecce925b06.png?imageMogr2/auto-orient/strip%7CimageView2/2/w/1240) 最近总结了一下数据结构和算法的题目,这是第二篇文章,关于链表的,第一篇文章关于二叉树的[参见](https://zhuanlan.zhihu.com/p/61445865) 废话少说,上**链表的数据结构** ``` class ListNode { ListNode next; int val; ListNode(int x){ val = x; next = null; } } ``` #### 1.翻转链表 ``` ListNode reverse(ListNode node){ ListNode prev = null; while(node!=null){ ListNode tmp = node.next; node.next = prev; prev = node; node = tmp; } return prev; } //翻转链表(递归方式) ListNode reverse2(ListNode head){ if(head.next == null){ return head; } ListNode reverseNode = reverse2(head.next); head.next.next = head; head.next = null; return reverseNode; } ``` #### 2.判断链表是否有环 ``` boolean hasCycle(ListNode head){ if(head == null|| head.next == null){ return false; } ListNode slow,fast; fast = head.next; slow = head; while(fast!=slow){ if(fast==null||fast.next==null){ return false; } fast = fast.next.next; slow = slow.next; } return true; } ``` #### 3,链表排序 ``` ListNode sortList(ListNode head){ if(head == null|| head.next == null){ return head; } ListNode mid = middleNode(head); ListNode right = sortList(mid.next); mid.next = null; ListNode left = sortList(head); return merge(left, right); } ListNode middleNode(ListNode head){ ListNode slow = head; ListNode fast = head.next; while(fast!=null&fast.next!=null){ slow = slow.next; fast = fast.next.next; } return slow; } ListNode merge(ListNode n1,ListNode n2){ ListNode dummy = new ListNode(0); ListNode node = dummy; while (n1!=null&&n2!=null) { if(n1.val<n2.val){ node.next = n1; n1 = n1.next; }else{ node.next = n2; n2 = n2.next; } node = node.next; } if(n1!=null){ node.next = n1; }else{ node.next = n2; } return dummy.next; } ``` #### 4.链表相加求和 ``` ListNode addLists(ListNode l1,ListNode l2){ if(l1==null&&l2==null){ return null; } ListNode head = new ListNode(); ListNode point = head; int carry = 0; while(l1!=null&&l2!=null){ int sum = carry + l1.val + l2.val; point.next = new ListNode(sum%10); carry = sum/10; l1 = l1.next; l2 = l2.next; point = point.next; } while(l1!=null){ int sum = carry + l1.val; point.next = new ListNode(sum%10); carry = sum/10; l1 = l1.next; point = point.next; } while(l2!=null){ int sum = carry + l2.val; point.next = new ListNode(sum%10); carry = sum/10; l2 = l2.next; point = point.next; } if(carry!=0){ point.next = new ListNode(carry); } return head.next; } ``` #### 5.得到链表倒数第n个节点 ``` ListNode nthToLast(ListNode head,int n ){ if(head == null||n<1){ return null; } ListNode l1 = head; ListNode l2 = head; for(int i = 0;i<n-1;i++){ if(l2 == null){ return null; } l2 = l2.next; } while(l2.next!=null){ l1 = l1.next; l2 = l2.next; } return l1; } ``` #### 6.删除链表倒数第n个节点 ``` ListNode deletenthNode(ListNode head,int n){ // write your code here if (n <= 0) { return null; } ListNode dumy = new ListNode(0); dumy.next = head; ListNode prdDel = dumy; for(int i = 0;i<n;i++){ if(head==null){ return null; } head = head.next; } while(head!=null){ head = head.next; prdDel = prdDel.next; } prdDel.next = prdDel.next.next; return dumy.next; } ``` #### 7.删除链表中重复的元素 ``` ListNode deleteMuNode(ListNode head){ if(head == null){ return null; } ListNode node = head; while(node.next != null){ if(node.val == node.next.val){ node.next = node.next.next; }else{ node = node.next; } } return head; } ``` #### 8.删除链表中重复的元素ii,去掉重复的节点 ``` ListNode deleteMuNode2(ListNode head){ if(head == null||head.next == null){ return head; } ListNode dummy = new ListNode(0); dummy.next = head; head = dummy; while(head.next!=null&&head.next.next!=null){ if(head.next.val == head.next.next.val){ int val = head.next.val; while(head.next.val == val&&head.next != null){ head.next = head.next.next; } }else{ head = head.next; } } return dummy.next; } ``` #### 9.旋转链表 ``` ListNode rotateRight(ListNode head,int k){ if(head ==null){ return null; } int length = getLength(head); k = k % length; ListNode dummy = new ListNode(0); dummy.next = head; head = dummy; ListNode tail = dummy; for(int i = 0;i<k;i++){ head = head.next; } while(head.next!= null){ head = head.next; tail = tail.next; } head.next = dummy.next; dummy.next = tail.next; tail.next = null; return dummy.next; } ``` #### 10.重排链表 ``` ListNode reOrder(ListNode head){ if(head == null||head.next == null){ return; } ListNode mid = middleNode(head); ListNode tail = reverse(mid.next); mergeIndex(head, tail); } private void mergeIndex(ListNode head1,ListNode head2){ int index = 0; ListNode dummy = new ListNode(0); while (head1!=null&&head2!=null) { if(index%2==0){ dummy.next = head1; head1 = head1.next; }else{ dummy.next = head2; head2 = head2.next; } dummy = dummy.next; index ++; } if(head1!=null){ dummy.next = head1; }else{ dummy.next = head2; } } ``` #### 11.链表划分 ``` ListNode partition(ListNode head,int x){ if(head == null){ return null; } ListNode left = new ListNode(0); ListNode right = new ListNode(0); ListNode leftDummy = left; ListNode rightDummy = right; while(head!=null){ if(head.val<x){ left.next = head; left = head; }else{ right.next = head; right = head; } head = head.next; } left.next = rightDummy.next; right.next = null; return leftDummy.next; } ``` #### 12.翻转链表的n到m之间的节点 ``` ListNode reverseN2M(ListNode head,int m,int n){ if(m>=n||head == null){ return head; } ListNode dummy = new ListNode(0); dummy.next = head; head = dummy; for(int i = 1;i<m;i++){ if(head == null){ return null; } head = head.next; } ListNode pmNode = head; ListNode mNode = head.next; ListNode nNode = mNode; ListNode pnNode = mNode.next; for(int i = m;i<n;i++){ if(pnNode == null){ return null; } ListNode tmp = pnNode.next; pnNode.next = nNode; nNode = pnNode; pnNode = tmp; } pmNode.next = nNode; mNode.next = pnNode; return dummy.next; } ``` #### 13.合并K个排序过的链表 ``` ListNode mergeKListNode(ArrayList<ListNode> k){ if(k.size()==0){ return null; } return mergeHelper(k,0,k.size()-1); } ListNode mergeHelper(List<ListNode> lists,int start,int end){ if(start == end){ return lists.get(start); } int mid = start + ( end - start )/2; ListNode left = mergeHelper(lists, start, mid); ListNode right = mergeHelper(lists, mid+1, end); return mergeTwoLists(left,right); } ListNode mergeTwoLists(ListNode list1,ListNode list2){ ListNode dummy = new ListNode(0); ListNode tail = dummy; while(list1!=null&&list2!=null){ if(list1.val<list2.val){ tail.next = list1; tail = tail.next; list1 = list1.next; }else{ tail.next = list2; tail = list2; list2 = list2.next; } } if(list1!=null){ tail.next = list1; }else{ tail.next = list2; } return dummy.next; } ``` #### 最后 **欢迎工作一到五年的Java工程师朋友们加入Java架构开发:277763288** **群内提供免费的Java架构学习资料(里面有高可用、高并发、高性能及分布式、Jvm性能调优、Spring源码,MyBatis,Netty,Redis,Kafka,Mysql,Zookeeper,Tomcat,Docker,Dubbo,Nginx等多个知识点的架构资料)合理利用自己每一分每一秒的时间来学习提升自己,不要再用"没有时间“来掩饰自己思想上的懒惰!趁年轻,使劲拼,给未来的自己一个交代!**

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