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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.
题意：
给出一个链表，删除所有重复的结点。
步骤：
（1）定义一个新的头结点，用来指向head，并将head向前移动一位。
1 -> 1 -> 1 -> 2 -> 3
变为 head -> 1 -> 1 -> 1 -> 2 -> 3
（2）遍历各结点，每次比较两个结点，head.next和head.next.next.
1. 如果head.next.val == head.next.next.val, 两个结点的值相等，用temp记录这个值，
从next开始依次遍历删除每个值为temp的结点，head.next = head.next.next.
2. 如果next和next.next的值不相等，继续向后遍历，head =head.next;
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode deleteDuplicates(ListNode head) { if(head == null) return null; 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 temp = head.next.val; while(head.next != null && head.next.val == temp){ head.next = head.next.next; } }else{ head = head.next; } } return dummy.next; } }
leetcode 82. Remove Duplicates from Sorted List II
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原文地址：http://www.cnblogs.com/iwangzheng/p/5714807.html
