LeetCode Remove Duplicates from Sorted List II

来源：互联网发布：喀秋莎软件的操作介绍编辑：程序博客网时间：2024/05/17 01:19

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.

题意：删除重复的所的节点。

思路：如果当前的节点的值=下一个节点的值的话就一直删掉，注意细节，每次我们都只是确定了一个节点，所以pre.next每次都要置null。

/** * 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 head;    ListNode cur = head;    ListNode root = new ListNode(Integer.MAX_VALUE);    ListNode pre = root;        while (cur != null && cur.next != null) {    if (cur.val == cur.next.val) {    int v = cur.val;    while (cur.val == v) {    cur = cur.next;    if (cur == null) return root.next;    }    } else {    pre.next = cur;    pre = pre.next;    cur = cur.next;    pre.next = null;    }    }        pre.next = cur;    return root.next;    }}

0 0