[数据结构与算法]Leetcode——删除排序链表中的重复元素

1. 删除排序链表中的重复元素1

题解

/**
 * 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) {
        ListNode 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;
    }
}

2. 删除排序链表中的重复元素2

题解（迭代）

/**
 * 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; }
 * }
 */
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return head;
        }
        ListNode node = new ListNode(0);
        node.next = head;
        ListNode cur = node;
        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 node.next;
    }
}

题解（递归）

思路：删除所有头部的重复节点，返回不重复的第一个节点。

1.特殊情况，头节点为null或头节点下一节点为null，直接返回头节点，这时不存在重复节点
2.如果头节点与，头节点的下一节点值相等，跳过所有相等节点。递归调用函数判断最后一个跳过节点的后一节点。
3.如果头节点与，头节点的下一节点值不等，递归调用函数判断头节点的后一节点。

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