
难度:Miduem
解题思路:是系列I的延伸,比较cur->next和cur->next->next,找到重复的地方,然后往后一直数,直到不重复。
代码如下:
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* deleteDuplicates(ListNode* head) { ListNode *dummy = new ListNode(-1), *current = dummy; dummy->next = head; while(current->next != NULL && current->next->next != NULL) { ListNode *current_n = current->next; ListNode *current_n_n = current_n->next; if(current_n ->val == current_n_n->val) { int value = current_n_n->val; while(current_n_n->next != NULL && current_n_n->next->val == value) { current_n_n = current_n_n->next; } current->next = current_n_n->next; } else { current = current->next; } } return dummy->next; } };
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代码结果:9ms,超过20.26%。
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