82. 删除排序链表中的重复元素 II

82. 删除排序链表中的重复元素 II

• 图解
  

class Solution {  //java 
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy;
        while (head != null && head.next != null) {
            if (head.val == head.next.val) {
                while (head.next != null && head.val == head.next.val) {
                    head = head.next;  //head移到一段中最后一个重复元素
                }
                pre.next = head.next;
            } else {
                pre = head;  //pre始终 在head左边
            }
            head = head.next;
        }
        return dummy.next;
    }
}

• 图解：太神了，此题的算法！
  
  
• 对比 eetcode 83题，利用外层while()，实现 cur.next = cur.next.next; 因为cur指针指向 while 的起始重复点。

class Solution {  // java 太神了
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode cur = dummy;
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                int x = cur.next.val;
                while (cur.next != null && cur.next.val == x) {
                    cur.next =cur.next.next;
                }
            } else {
               cur = cur.next;
            }
        }
        return dummy.next;
    }
}