【Leetcode】Remove Duplicates from Sorted List II

题目链接：https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

题目：

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

思路：

思路很简单，唯一要注意的是 删除操作加入头结点可以减少很多判断。

算法：

public ListNode deleteDuplicates(ListNode head) {if(head==null)return null;ListNode newHead = new ListNode(0);newHead.next = head;ListNode p = head, q, pre = newHead;// pre p qq = p.next;while (q != null) {if (p.val == q.val) {while (q != null && p.val == q.val) {// 如果节点相等则前移至后面第一个不等节点q = q.next;}pre.next = q;p = pre;//记得把p要更新，否则p指向的是要删除的结点,下次循环进入else会出错。p不能更新为q，否则下轮循环p/q相等进入死循环。} else {pre = p;p = q;q = q.next;}}return newHead.next;}

算法2：没加入头结点，简直超麻烦，判断边界条件神烦啊。。自从删除、增加结点操作加入头结点之后，代码简洁可读性强多了。亲身体会！

public ListNode deleteDuplicates(ListNode head) {ListNode p = head, q, pre = null;if (p == null) {return head;}// pre p qq = p.next;while (q != null) {if (p.val == q.val) {while (q != null && p.val == q.val) {// 如果节点相等则前移至第一个不等节点q = q.next;}if (pre == null) {// 如果头节点重复,需要特殊处理head = q;// 如此时1,1,2,2p = q;// head、p都指向2if (q != null) // 若不是1,1这种情况，即q为null时，q向前移到第二个2位置q = q.next;} else { //pre.next = q;p = pre;}} else {pre = p;p = q;q = q.next;}}return head;}