LeetCode题解-82-Remove Duplicates from Sorted List II
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原题
解题思路
利用两个节点node1与node2进行遍历,node2是node1的后驱节点
if(node1.next.val == node2.next.val) 那么此时存在重复的节点,
记录下此值val,将node1之后的所有值为val的节点删除;
else
node1与node2前进1步;
通过一次遍历可以将链表中重复的元素删除完毕。
图解
代码
public class Solution82 { public ListNode deleteDuplicates(ListNode head) { if(head == null || head.next == null) return head; ListNode dummy = new ListNode(0); dummy.next = head; ListNode node1 = dummy; ListNode node2 = dummy.next; while (node2 != null && node2.next != null){ if (node1.next.val == node2.next.val){ int val = node1.next.val; while (node1.next != null && node1.next.val == val){ node1.next = node2.next; node2 = node1.next; } } else { node1 = node2; node2 = node1.next; } } return dummy.next; }}
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