83. Remove Duplicates from Sorted List && 82. Remove Duplicates from Sorted List II

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#数据结构与算法

本文介绍两种场景下删除链表中重复元素的方法：一种是删除所有重复出现的元素，只保留唯一值；另一种是删除所有连续重复的元素。通过示例展示了不同情况下的链表操作，并提供了完整的Java代码实现。

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Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

  public ListNode deleteDuplicates(ListNode head) {
       if(head == null || head.next == null)
            return head;
       ListNode p = head,q = head.next;     //p记录无重复的链表，q记录处理的下一个节点

        while(q != null){
            if(p.val != q.val){  //没有重复
                p.next = q;
                p = q;
            }
            q = q.next;

        }
        p.next = q;
        return head;
    }

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

 public ListNode deleteDuplicates(ListNode head) {
        if(head == null) return head;

        ListNode fakehead = new ListNode(0), cur = head,pre = fakehead;
        fakehead.next = head;
        while(cur != null && cur.next != null){   
            if(cur.next.val == cur.val){                  //去重复
                ListNode t = cur.next.next;
                while(t != null && t.val == cur.val){
                    t = t.next;
                }
                cur = t;
            }
            else{                                    //没有重复
                pre.next = cur;
                pre = cur;
                cur = cur.next;
            }

        }
        pre.next = cur;
        return fakehead.next;
    }