LeetCode[Linked List]: Remove Duplicates from Sorted List II

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#LeetCode #Linked List #链表 #算法 #虚拟头结点

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本文介绍了一种从已排序链表中移除所有具有重复数值的节点的方法，只保留原始列表中的唯一数值。通过使用虚拟头结点技术来简化处理流程。

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.

很常规的解法，为去除头节点的特殊性，需要用到虚拟头结点技术。

    ListNode *deleteDuplicates(ListNode *head) {
        if (!head) return NULL;

        ListNode *dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode *prev = dummyHead, *curr = head->next;
        int curVal = head->val;
        while (curr) {
            if (curr->val == curVal) {
                for (curr = curr->next; curr != NULL && curr->val == curVal; curr = curr->next) ;
                prev->next = curr;
                if (curr) {
                    curVal = curr->val;
                    curr = curr->next;
                }
            }
            else {
                prev = prev->next;
                curVal = curr->val;
                curr = curr->next;
            }
        }

        return dummyHead->next;
    }