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解题思路

思路一: 迭代 快慢指针

用快指针跳过那些有重复数组,慢指针负责和快指针拼接!

  1. public ListNode deleteDuplicates(ListNode head) {
  2.         if (head == null)
  3.             return head;
  4.         ListNode dummy = new ListNode(-1);
  5.         dummy.next = head;
  6.         ListNode slow = dummy;
  7.         ListNode fast = dummy.next;
  8.         while (fast != null){
  9.             while (fast.next != null && fast.val == fast.next.val)
  10.                 fast = fast.next;
  11.             if (slow.next == fast)
  12.                 slow = slow.next;
  13.             else
  14.                 slow.next = fast.next;
  15.             fast = fast.next;
  16.         }
  17.         return dummy.next;
  18.     }