原地删除有序链表中的重复节点

用哑节点 + 快慢指针

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode deleteDuplicates (ListNode head) {
        // write code here
        if(head == null) return null;
        //定义哑节点
        ListNode newNode = new ListNode(-1);
        newNode.next = head;
        //快慢指针
        ListNode slow = newNode, fast = head;

        while(fast != null && fast.next != null){
            //若前后值相等
            if(fast.val == fast.next.val){
                //就循环找到一个不相等的节点
                ListNode temp = fast.next;
                while(temp != null && temp.val == fast.val){
                    temp = temp.next;
                }
                //慢指针的下一个是只出现过一次的
                slow.next = temp;
                fast = temp;
            }else{
                //不相等，则前进
                slow = slow.next;
                fast = fast.next;
            }

        }
        return newNode.next;
    }
}

注意此题是只保留出现一次的元素。
若只删除重复出现的元素，比如：

同样是用快慢指针，但不需要用到哑节点

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        //利用快慢指针
        if(head == null)
            return null;
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast != null){
            if(slow.val != fast.val){
                slow.next = fast;
                slow = slow.next;
            }
            fast = fast.next;
        }
        slow.next = null;
        return head;
    }
}