Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given1->2->3->3->4->4->5
, return1->2->5
.
Given1->1->1->2->3
, return2->3
.
非经常规的解法。为去除头节点的特殊性,须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; }