[LeetCode] 83. Remove Duplicates from Sorted List

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Example 1:

Input: head = [1,1,2]
Output: [1,2]

Example 2:

Input: head = [1,1,2,3,3]
Output: [1,2,3]

Constraints:
The number of nodes in the list is in the range [0, 300].
-100 <= Node.val <= 100
The list is guaranteed to be sorted in ascending order.

删除排序链表中的重复元素。

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 。

思路

思路很简单，每次去看一下当前节点和下一个节点的val是否相同，如是，则跳过下一个节点。

复杂度

时间O(n)
空间O(1)

代码

Java实现

/**
 * 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) {
        // corner case
        if (head == null || head.next == null) {
            return head;
        }

        // normal case
        ListNode cur = head;
        while (cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }
}

JavaScript实现

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteDuplicates = function(head) {
    // corner case
    if (head === null || head.next === null) {
        return head;
    }

    // normal case
    let 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;
};

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