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题目:
思路:
双指针,slow和fast,并且增加标记flag初始为1。
如果slow指向节点值等于fast指向节点值,fast向后走,flag置为0;
如果slow指向节点值不等于fast指向节点值,观察flag的值若为0,slow指向fast,fast向后走,flag置为1,然后continue;观察flag的值若不为0,将该节点拿下来,成为我们的目标节点去处理。
剩下的就是细节以及最后一个节点的问题,比较简单,判断一下就好。
代码:
struct ListNode* deleteDuplicates(struct ListNode* head)
{// write code hereif (head == NULL || head->next == NULL)return head;struct ListNode* tail = NULL;struct ListNode* newhead = NULL;struct ListNode* slow = head;struct ListNode* fast = slow->next;int flag = 1;while (fast){if (slow->val == fast->val){fast = fast->next;flag = 0;}else{if (flag == 0){slow = fast;fast = fast->next;flag = 1;continue;}if (newhead == NULL){tail = newhead = slow;}else{tail->next = slow;tail = slow;}slow = fast;fast = fast->next;}}if (flag == 1){if (tail)tail->next = slow;elsenewhead = slow;}else{if (tail)tail->next = NULL;}return newhead;
}
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