[LeetCode] 83. 删除排序链表中的重复元素

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

题目描述:

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

示例:

示例 1:

输入: 1->1->2
输出: 1->2

示例 2:

输入: 1->1->2->3->3
输出: 1->2->3

思路:

思路一:迭代,快慢指针,更容易理解

思路二:递归

自己看代码,很好理解!

相关题目:82. 删除排序链表中的重复元素 II

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:
        dummy = ListNode(-1000)
        dummy.next = head
        slow = dummy
        fast = dummy.next
        while fast :
            if slow.val == fast.val:
                fast = fast.next
                slow.next = fast
            else:
                slow = slow.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) {
        ListNode dummy = new ListNode(-1000);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy.next;
        while (fast != null) {
            if (slow.val == fast.val) {
                fast = fast.next;
                slow.next = fast;
            } else {
                slow = slow.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)
        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);
        }
        else head.next = deleteDuplicates(head.next);
        return head;    
    }
}