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100 lines (79 loc) · 2.52 KB

存在一个按升序排列的链表,给你这个链表的头节点 head ,请你删除所有重复的元素,使每个元素 只出现一次 。

返回同样按升序排列的结果链表。

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

实现一

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

        // 这种删除节点的常用方式就是给一个哑节点
        ListNode dummyNode = new ListNode(0, head);
        ListNode cur = dummyNode;
        while (null != cur.next && null != cur.next.next) {
            if (cur.next.val == cur.next.next.val) {
                while (null != cur.next && null != cur.next.next && cur.next.val == cur.next.next.val) {
                    cur.next = cur.next.next;
                }
            } else {
                cur = cur.next;
            }
        }

        return dummyNode.next;
    }
}

实现二

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (null == head || null == head.next) {
            return head;
        }

        // 不创建哑节点
        ListNode cur = head;
        while (null != cur.next) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }

        return head;
    }
}

实现三

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (null == head || null == head.next) {
            return head;
        }
        // 递归
        if (head.val == head.next.val) {
            ListNode node = head.next;
            head.next = head.next.next;
            node.next = null;
            deleteDuplicates(head);
        } else {
            deleteDuplicates(head.next);
        }

        return head;
    }
}