[LeetCode] 82. 删除排序链表中的重复元素 II

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

题目描述:

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例:

示例 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5

示例 2:

输入: 1->1->1->2->3
输出: 2->3

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.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        dummy = ListNode(-1000)
        dummy.next = head
        slow = dummy
        fast = dummy.next
        while fast:
            if  fast.next and fast.next.val == fast.val:
                tmp = fast.val
                while fast and tmp == fast.val:
                    fast = fast.next
            else:
                slow.next = fast
                slow = fast
                fast = fast.next
        slow.next = fast
        return dummy.next

思路一(另一个版本):

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def deleteDuplicates(self, head):
        # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        dummy = ListNode(-1)
        dummy.next = head
        slow = dummy
        fast = dummy.next
        while fast:
            while fast.next and slow.next.val == fast.next.val:
                fast = fast.next
            if slow.next == fast:
                slow = fast
            else:
                slow.next = fast.next
            fast = fast.next
        return dummy.next

java

/**
 * 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.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if not head:return head
        if head.next and head.val == head.next.val:
            while head.next != None and head.val == head.next.val:
                head = head.next
            return self.deleteDuplicates(head.next)
        else:
            head.next = self.deleteDuplicates(head.next)
        return head

java

/**
 * 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;
    }
}