javascript 解法, 执行用时: 64 ms, 内存消耗: 50.2 MB, 提交时间: 2024-01-14 12:19:55

/**
 * 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;

    let cur = head;
    while (cur.next) {
        if (cur.val === cur.next.val) {
            cur.next = cur.next.next;
        } else {
            cur = cur.next;
        }
    }
    return head;
};

java 解法, 执行用时: 0 ms, 内存消耗: 43.1 MB, 提交时间: 2024-01-14 12:19:31

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

        return head;
    }
}

cpp 解法, 执行用时: 8 ms, 内存消耗: 11.8 MB, 提交时间: 2024-01-14 12:18:58

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

        return head;
    }
};

golang 解法, 执行用时: 4 ms, 内存消耗: 3.1 MB, 提交时间: 2021-07-20 09:58:26

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

golang 解法, 执行用时: 4 ms, 内存消耗: 3.6 MB, 提交时间: 2021-07-20 09:57:34

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteDuplicates(head *ListNode) *ListNode {
	sentinel := &ListNode{Val:0}
	sentinel.Next = head
	curr, prev := head, sentinel
    mp := map[int]int{}
	for curr != nil {
		if _, ok := mp[curr.Val]; ok {
			prev.Next = curr.Next
		} else {
			prev = curr
            mp[curr.Val] = 1
		}
		curr = curr.Next
	}
	return sentinel.Next
}

python3 解法, 执行用时: 48 ms, 内存消耗: 13.6 MB, 提交时间: 2020-11-16 18:21:53

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        head.next = self.deleteDuplicates(head.next)
        if head.val == head.next.val:
            head = head.next
        return head