82. 删除排序链表中的重复元素 II 

题目描述

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

解法

方法一：一次遍历

我们先创建一个虚拟头节点 $dummy$，令 $dummy.next = head$，然后创建指针 $pre$ 指向 $dummy$，指针 $cur$ 指向 $head$，开始遍历链表。

当 $cur$ 指向的节点值与 $cur.next$ 指向的节点值相同时，我们就让 $cur$ 不断向后移动，直到 $cur$ 指向的节点值与 $cur.next$ 指向的节点值不相同时，停止移动。此时，我们判断 $pre.next$ 是否等于 $cur$，如果相等，说明 $pre$ 与 $cur$ 之间没有重复节点，我们就让 $pre$ 移动到 $cur$ 的位置；否则，说明 $pre$ 与 $cur$ 之间有重复节点，我们就让 $pre.next$ 指向 $cur.next$。然后让 $cur$ 继续向后移动。继续上述操作，直到 $cur$ 为空，遍历结束。

最后，返回 $dummy.next$ 即可。

时间复杂度 $O(n)$，其中 $n$ 为链表的长度。空间复杂度 $O(1)$。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = pre = ListNode(next=head)
        cur = head
        while cur:
            while cur.next and cur.next.val == cur.val:
                cur = cur.next
            if pre.next == cur:
                pre = cur
            else:
                pre.next = cur.next
            cur = cur.next
        return dummy.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy;
        ListNode cur = head;
        while (cur != null) {
            while (cur.next != null && cur.next.val == cur.val) {
                cur = cur.next;
            }
            if (pre.next == cur) {
                pre = cur;
            } else {
                pre.next = cur.next;
            }
            cur = cur.next;
        }
        return dummy.next;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* dummy = new ListNode(0, head);
        ListNode* pre = dummy;
        ListNode* cur = head;
        while (cur) {
            while (cur->next && cur->next->val == cur->val) {
                cur = cur->next;
            }
            if (pre->next == cur) {
                pre = cur;
            } else {
                pre->next = cur->next;
            }
            cur = cur->next;
        }
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteDuplicates(head *ListNode) *ListNode {
	dummy := &ListNode{Next: head}
	pre, cur := dummy, head
	for cur != nil {
		for cur.Next != nil && cur.Next.Val == cur.Val {
			cur = cur.Next
		}
		if pre.Next == cur {
			pre = cur
		} else {
			pre.Next = cur.Next
		}
		cur = cur.Next
	}
	return dummy.Next
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function deleteDuplicates(head: ListNode | null): ListNode | null {
    const dummy = new ListNode(0, head);
    let pre = dummy;
    let cur = head;
    while (cur) {
        while (cur.next && cur.val === cur.next.val) {
            cur = cur.next;
        }
        if (pre.next === cur) {
            pre = cur;
        } else {
            pre.next = cur.next;
        }
        cur = cur.next;
    }
    return dummy.next;
}

Rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn delete_duplicates(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode::new(101)));
        let mut pev = dummy.as_mut().unwrap();
        let mut cur = head;
        let mut pre = 101;
        while let Some(mut node) = cur {
            cur = node.next.take();
            if node.val == pre || (cur.is_some() && cur.as_ref().unwrap().val == node.val) {
                pre = node.val;
            } else {
                pre = node.val;
                pev.next = Some(node);
                pev = pev.next.as_mut().unwrap();
            }
        }
        dummy.unwrap().next
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteDuplicates = function (head) {
    const dummy = new ListNode(0, head);
    let pre = dummy;
    let cur = head;
    while (cur) {
        while (cur.next && cur.val === cur.next.val) {
            cur = cur.next;
        }
        if (pre.next === cur) {
            pre = cur;
        } else {
            pre.next = cur.next;
        }
        cur = cur.next;
    }
    return dummy.next;
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode DeleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy;
        ListNode cur = head;
        while (cur != null) {
            while (cur.next != null && cur.next.val == cur.val) {
                cur = cur.next;
            }
            if (pre.next == cur) {
                pre = cur;
            } else {
                pre.next = cur.next;
            }
            cur = cur.next;
        }
        return dummy.next;
    }
}