Leetcode: Remove Duplicates from Sorted List II 解题报告

Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

SOLUTION 1:

使用一个del标记来删除最后一个重复的字元。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return null;
        }
        
        // record the head.
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode cur = dummy;
        
        // to delete the last node in the list of duplications.
        boolean del = false;
        
        while (cur != null) {
            if (cur.next != null
                && cur.next.next != null
                && cur.next.val == cur.next.next.val) {
                cur.next = cur.next.next;
                del = true;
            } else {
                if (del) {
                    cur.next = cur.next.next;
                    del = false;
                } else {
                    cur = cur.next;
                }
            }
        }
        
        return dummy.next;
    }
}

SOLUTION 2:

使用一个pre, 一个cur来扫描，遇到重复的时候，使用for循环用cur跳过所有重复的元素。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return null;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode pre = dummy;
        ListNode cur = pre.next;
        
        while (cur != null && cur.next != null) {
            if (cur.val == cur.next.val) {
                while (cur != null && cur.val == pre.next.val) {
                    cur = cur.next;
                }
                
                // delete all the duplication.
                pre.next = cur;
            } else {
                cur = cur.next;
                pre = pre.next;
            }
        }
        
        return dummy.next;
    }
}