从0开始的秋招刷题路，记录下所刷每道题的题解，帮助自己回顾总结

82. 删除排序链表中的重复元素 II

给定一个已排序的链表的头 head ， 删除原始链表中所有重复数字的节点，只留下不同的数字 。返回 已排序的链表 。

示例 1：

输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]

示例 2：

输入：head = [1,1,1,2,3]
输出：[2,3]

提示：
链表中节点数目在范围 [0, 300] 内
-100 <= Node.val <= 100
题目数据保证链表已经按升序 排列

思路:

思路一: 迭代 快慢指针,用快指针跳过那些有重复数组,慢指针负责和快指针拼接!

思路二:递归

相关题目:83. 删除排序链表中的重复元素

代码:

思路一:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return head;
        ListNode dummy = new ListNode(-1000);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy.next;
        while (fast != null) {
            while (fast.next != null && fast.val == fast.next.val) fast = fast.next;
            if (slow.next == fast) slow = slow.next;
            else slow.next = fast.next;
            fast = fast.next;
        }
        return dummy.next; 
    }
}

思路二:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null)  return head;
        if (head.next != null && head.val == head.next.val) {
            while (head.next != null && head.val == head.next.val) {
                head = head.next;
            }
            return deleteDuplicates(head.next);
        }
        else head.next = deleteDuplicates(head.next);
        return head;    
    }
}