1.反转一个单链表

https://leetcode-cn.com/problems/reverse-linked-list/description/

 public ListNode reverseList(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode prev = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode curNext = cur.next;
            cur.next = prev;
            prev = cur;
            cur = curNext;
        }
        return prev;
    }

2. 链表的中心结点

https://leetcode-cn.com/problems/middle-of-the-linked-list/description/

  public ListNode middleNode(ListNode head) {
        if (head == null) {
            return null;
        }
        if (head.next == null) {
            return head;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

3 链表中倒数第k个结点

https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&&tqId=11167&rp=2&ru=/activity/oj&qru=/ta/coding-interviews/question-ranking

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        //k是否是合法的
        if(k <= 0 || head == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
         for (int i = 0; i < k; i++) {
            if (fast ==null) {
                return null;
            }
            fast = fast.next;
        }
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

4 合并两个有序链表 

 ​​​​​​https://leetcode-cn.com/problems/merge-two-sorted-lists/

class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode newNode = new ListNode(0);//虚拟节点，数据不具备意义。
        ListNode tmp = newNode;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                tmp.next = list1;
                tmp = tmp.next;
                list1 = list1.next;
            }else {
                tmp.next = list2;
                tmp = tmp.next;
                list2 = list2.next;
            }
        }
        if(list1 == null && list2 != null) {
            tmp.next = list2;
        } else {
            tmp.next = list1;
        }
        return newNode.next;
    }
}

5 链表分割

https://www.nowcoder.com/practice/0e27e0b064de4eacac178676ef9c9d70?tpId=8&&tqId=11004&rp=2&ru=/activity/oj&qru=/ta/cracking-the-coding-interview/question-ranking

public class Partition {
    public ListNode partition(ListNode pHead, int x) {
        // write code here
  if (pHead == null) {
            return null;
        }
        ListNode bs = null;
        ListNode be = null;
        ListNode as = null;
        ListNode ae = null;

        ListNode cur = pHead;
        while (cur != null) {
            if (cur.val < x) {
                if (bs == null) {
                    bs = cur;
                    be = cur;
                } else {
                    be.next = cur;
                    be = be.next;
                }
            } else {
                if (as == null) {
                    as = cur;
                    ae = cur;
                } else {
                    ae.next = cur;
                    ae = ae.next;
                }
            }
            cur = cur.next;
        }

        if (bs == null) {
            return as;
        }
        be.next = as;
        if (as != null) {
            ae.next = null;
        }
        return bs;
    }
}