删除排序链表中的重复元素 II

82. 删除排序链表中的重复元素 II

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

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删除排序链表中的重复元素

原站题解

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上次编辑到这里，代码来自缓存点击恢复默认模板

/** * 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) { } };

javascript 解法, 执行用时: 72 ms, 内存消耗: 50.2 MB, 提交时间: 2024-01-15 00:04:49

/**
 * 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) {
    if (!head) {
        return head;
    }

    const dummy = new ListNode(0, head);

    let cur = dummy;
    while (cur.next && cur.next.next) {
        if (cur.next.val === cur.next.next.val) {
            const x = cur.next.val;
            while (cur.next && cur.next.val === x) {
                cur.next = cur.next.next;
            } 
        } else {
            cur = cur.next;
        }
    }
    return dummy.next;
};

golang 解法, 执行用时: 4 ms, 内存消耗: 2.7 MB, 提交时间: 2024-01-15 00:04:23

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

    dummy := &ListNode{0, head}

    cur := dummy
    for cur.Next != nil && cur.Next.Next != nil {
        if cur.Next.Val == cur.Next.Next.Val {
            x := cur.Next.Val
            for cur.Next != nil && cur.Next.Val == x {
                cur.Next = cur.Next.Next
            }
        } else {
            cur = cur.Next
        }
    }

    return dummy.Next
}

java 解法, 执行用时: 0 ms, 内存消耗: 41.7 MB, 提交时间: 2024-01-15 00:04:04

/**
 * 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) {
        if (head == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0, head);

        ListNode cur = dummy;
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                int x = cur.next.val;
                while (cur.next != null && cur.next.val == x) {
                    cur.next = cur.next.next;
                }
            } else {
                cur = cur.next;
            }
        }

        return dummy.next;
    }
}

cpp 解法, 执行用时: 12 ms, 内存消耗: 11.2 MB, 提交时间: 2024-01-15 00:03:45

/**
 * 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) {
        if (!head) {
            return head;
        }
        
        ListNode* dummy = new ListNode(0, head);

        ListNode* cur = dummy;
        while (cur->next && cur->next->next) {
            if (cur->next->val == cur->next->next->val) {
                int x = cur->next->val;
                while (cur->next && cur->next->val == x) {
                    cur->next = cur->next->next;
                }
            }
            else {
                cur = cur->next;
            }
        }

        return dummy->next;
    }
};

python3 解法, 执行用时: 48 ms, 内存消耗: 15 MB, 提交时间: 2022-08-29 11:15:37

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
'''
一次遍历，
'''
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if not head:
            return head
        
        dummy = ListNode(0, head)

        cur = dummy
        while cur.next and cur.next.next:
            if cur.next.val == cur.next.next.val:
                x = cur.next.val
                while cur.next and cur.next.val == x:
                    cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy.next

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