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82. 删除排序链表中的重复元素 II
给定一个已排序的链表的头 head
, 删除原始链表中所有重复数字的节点,只留下不同的数字 。返回 已排序的链表。
示例 1:

输入:head = [1,2,3,3,4,4,5] 输出:[1,2,5]
示例 2:

输入:head = [1,1,1,2,3] 输出:[2,3]
提示:
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链表中节点数目在范围
[0, 300]
内 -
-100 <= Node.val <= 100
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题目数据保证链表已经按升序 排列
思路分析
快慢指针是个好办法!
注意推敲满指针前进的条件!
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一刷
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二刷
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三刷
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/**
* Runtime: 1 ms, faster than 60.61% of Java online submissions for Remove Duplicates from Sorted List II.
* Memory Usage: 39.4 MB, less than 6.98% of Java online submissions for Remove Duplicates from Sorted List II.
*
* Copy from: https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/solutions/7396/kuai-man-zhi-zhen-by-powcai-2/[递归与非递归 - 删除排序链表中的重复元素 II - 力扣(LeetCode)]
*/
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode slow = dummy;
ListNode fast = dummy;
while (Objects.nonNull(fast)) {
while (Objects.nonNull(fast.next) && fast.val == fast.next.val) {
fast = fast.next;
}
if (slow.next == fast) {
slow = slow.next;
} else {
slow.next = fast.next;
}
fast = fast.next;
}
return dummy.next;
}
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/**
* 参考官方题解
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-03 16:50:50
*/
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode();
dummy.next = head;
ListNode cur = dummy;
while (cur.next != null && cur.next.next != null) {
if (cur.next.val == cur.next.next.val) {
// 将重复元素全部删除
int val = cur.next.val;
while (cur.next != null && cur.next.val == val) {
cur.next = cur.next.next;
}
} else {
cur = cur.next;
}
}
return dummy.next;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-05-20 14:39:15
*/
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(119);
dummy.next = head;
ListNode pre = dummy;
ListNode slow = head;
ListNode fast = head;
boolean flag = false;
while (fast != null) {
if (fast.val == slow.val) {
if (fast != slow) {
flag = true;
}
fast = fast.next;
} else {
if (flag) {
pre.next = fast;
slow = fast;
flag = false;
} else {
pre = pre.next;
slow = slow.next;
fast = fast.next;
}
}
}
if (flag) {
pre.next = fast;
}
return dummy.next;
}