Question

83. 删除排序链表中的重复元素

English Version

题目描述

给定一个已排序的链表的头

 head ， _删除所有重复的元素，使每个元素只出现一次_ 。返回 _已排序的链表_ 。

 

示例 1：

输入：head = [1,1,2]
输出：[1,2]

示例 2：

输入：head = [1,1,2,3,3]
输出：[1,2,3]

 

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

我们用一个指针 $cur$ 来遍历链表。如果当前 $cur$ 与 $cur.next$ 对应的元素相同，我们就将 $cur$ 的 $next$ 指针指向 $cur$ 的下下个节点。否则，说明链表中 $cur$ 对应的元素是不重复的，因此可以将 $cur$ 指针移动到下一个节点。

遍历结束后，返回链表的头节点即可。

时间复杂度 $O(n)$，其中 $n$ 是链表的长度。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]:
    cur = head
    while cur and cur.next:
      if cur.val == cur.next.val:
        cur.next = cur.next.next
      else:
        cur = cur.next
    return head

/**
 * 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 cur = head;
    while (cur != null && cur.next != null) {
      if (cur.val == cur.next.val) {
        cur.next = cur.next.next;
      } else {
        cur = cur.next;
      }
    }
    return head;
  }
}

/**
 * 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* cur = head;
    while (cur != nullptr && cur->next != nullptr) {
      if (cur->val == cur->next->val) {
        cur->next = cur->next->next;
      } else {
        cur = cur->next;
      }
    }
    return head;
  }
};

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

// 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(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut dummy = Some(Box::new(ListNode::new(i32::MAX)));
    let mut p = &mut dummy;

    let mut current = head;
    while let Some(mut node) = current {
      current = node.next.take();
      if p.as_mut().unwrap().val != node.val {
        p.as_mut().unwrap().next = Some(node);
        p = &mut p.as_mut().unwrap().next;
      }
    }
    dummy.unwrap().next
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteDuplicates = function (head) {
  let cur = head;
  while (cur && cur.next) {
    if (cur.next.val === cur.val) {
      cur.next = cur.next.next;
    } else {
      cur = cur.next;
    }
  }
  return head;
};

/**
 * 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 cur = head;
    while (cur != null && cur.next != null) {
      if (cur.val == cur.next.val) {
        cur.next = cur.next.next;
      } else {
        cur = cur.next;
      }
    }
    return head;
  }
}