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.
这个与Remove Duplicates from Sorted List不同，本题是要删除所有重复的元素，即重复的就要删除，一个不剩，而后者是要删除重复的元素，只保留一个元素.
设p为当前的元素，初始化为head, pre为前一个元素，初始化为null， q为后一个元素，即q = p->next
- 若
q->val != p->val, 则直接更新p，pre
pre = p;
p = p->next;- 否则指向删除q，直到q与p值不相等
while(q && q->val == p->val) {
p->next = p->next->next;
free(q);
q = p->next;
}然后执行以下操作:
- 若
p == head, 即头节点就出现重复，需要更新head指针，并释放p
head = p->next;
free(p);
p = head;- 若
p != head, 则只需要删除p节点即可，利用pre指针
pre->next = pre->next->next;
free(p);
p = q;struct ListNode* deleteDuplicates(struct ListNode *head)
{
struct ListNode *p = head, *pre = NULL;
while (p) {
struct ListNode *q = p->next; // q is the next node of p
bool isDup = false;
while (q && q->val == p->val) {
isDup = true;
// delete q
p->next = p->next->next;
free(q);
// update q
q = p->next;
}
if (isDup) {
if (p == head) {
head = p->next;
free(p);
p = head;
} else {
pre->next = pre->next->next;
free(p);
p = q;
}
} else {
pre = p;
p = p->next;;
}
}
return head;
}Remove Duplicates from Sorted List I, 删除重复的元素，即重复的只保留一个，相当于去重