83. 删除排序链表中的重复元素

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 \(O(1)\)。

Python3JavaC++GoRustJavaScriptC#

# 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;
    }
}

83. 删除排序链表中的重复元素

题目描述

解法

方法一：一次遍历

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