Given a sorted linked list, delete all duplicates such that each element appear only once.
Example 1:
Input: 1->1->2
Output: 1->2
Example 2:
Input: 1->1->2->3->3
Output: 1->2->3
有序链表去重。
移除给定有序链表的重复项，那么我们可以遍历这个链表，每个结点和其后面的结点比较，如果结点值相同了，我们只要将前面结点的next指针跳过紧挨着的相同值的结点，指向后面一个结点。这样遍历下来，所有重复的结点都会被跳过，留下的链表就是没有重复项的了。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { //me public ListNode deleteDuplicates(ListNode head) { if (null == head) { return head; } ListNode cur = head; ListNode next = cur.next; while (cur != null && next != null) { while (next != null && cur.val == next.val) { cur.next = next.next; next = next.next; } cur = cur.next; } return head; } }
[LeetCode] 83. Remove Duplicates from Sorted List ☆(从有序数组中删除重复项)
原文：https://www.cnblogs.com/fanguangdexiaoyuer/p/10777216.html