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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:

0082 01
输入:head = [1,2,3,3,4,4,5]
输出:[1,2,5]

示例 2:

0082 02
输入:head = [1,1,1,2,3]
输出:[2,3]

提示:

  • 链表中节点数目在范围 [0, 300]

  • -100 <= Node.val <= 100

  • 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!

  • 一刷

  • 二刷

  • 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;
}