Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given, return
1->2->3->3->4->4->5.
1->2->5
Given, return
1->1->1->2->3.
2->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;
}