一篇文章搞定面试中的链表题目(java实现)

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;
    }