链表面试题代码总结（java）

这是原文链接：http://www.jianshu.com/p/a64d1ef95980

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
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;    }