Question

82. 删除排序链表中的重复元素 II

English Version

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 $O(1)$。

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

// 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
  }
}

/**
 * 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;
};

/**
 * 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;
  }
}