【经典算法】Leetcode.83删除排序链表中的重复元素(Java/C/Python3/Go实现含注释说明,Easy)

题目：删除排序链表中的重复元素

解法：遍历链表，并比较当前节点与下一节点的值，如果发现重复，则跳过下一节点并释放它。

Java 实现：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return head;
        }
 
        ListNode current = head;
        while (current.next != null) {
            if (current.val == current.next.val) {
                current.next = current.next.next;
            } else {
                current = current.next;
            }
        }
 
        return head;
    }
}

C 实现：

struct ListNode* deleteDuplicates(struct ListNode* head) {
    if (head == NULL) {
        return head;
    }
 
    struct ListNode* current = head;
    while (current->next != NULL) {
        if (current->val == current->next->val) {
            current->next = current->next->next;
        } else {
            current = current->next;
        }
    }
 
    return head;
}

Python3 实现：

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if not head:
            return head
 
        current = head
        while current.next:
            if current.val == current.next.val:
                current.next = current.next.next
            else:
                current = current.next
        return head

Go 实现：

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