LeetCode 82,83 Remove Duplicates from Sorted List I,II解析

/************************************************************************ * 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. ************************************************************************/

 //    1->2->1   1->2->1，               
 ListNode* deleteDuplicates(ListNode* head) {
        ListNode* cur = head;
        while(cur) {
            while(cur->next && cur->val == cur->next->val) {
                cur->next = cur->next->next;
            }
            cur = cur->next;
        }
        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.
************************************************************************////1->2->1は1->2->1を出力し、連続するノードの重複値のみを除去する必要があります
削除するのは最初のノードかもしれないので、最初のノードを指す偽のノードを見ると便利です.

ListNode *deleteDuplicates(ListNode *head) {
        ListNode preHead(0);
        preHead.next = head;
        ListNode* pre = &preHead;
        ListNode* cur = pre->next;
        while (cur) {
            ListNode* nxt = cur->next;
            //     
            if (nxt&& cur->val == nxt->val) {
                // move pn to next different value
                while (nxt && cur->val == nxt->val) {
                    nxt = nxt->next;
                }
                //    ，         pre  
                cur = nxt;
                pre->next = cur;
            }
            else {   //     
                pre = cur;     // pre cur,    
                cur = cur->next;
            }
        }
        return preHead.next;
    }