- Total Accepted: 163106
- Total Submissions: 417745
- Difficulty: Easy
- Contributors: Admin
Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2
, return 1->2
.
Given 1->1->2->3->3
, return 1->2->3
.
分析
使用双指针, pre 和 cur,非递归
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | /** * 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) { if (head == NULL || head->next == NULL) return head; ListNode* pre = head; ListNode* cur = head->next; while (cur != NULL){ if (cur->val == pre->val){ ListNode * tmp = cur; cur = cur->next; pre->next = cur; delete tmp; } else { pre = pre->next; cur = cur->next; } } return head; } }; |
递归法
删除head之后的list中重复元素,
然后再比较,如果head->val == head->next->val
则返回head->next
1 2 3 4 5 6 7 8 9 10 11 12 13 | class Solution { public : ListNode* deleteDuplicates(ListNode* head) { if (head == NULL || head->next == NULL) return head; head->next = deleteDuplicates(head->next); if (head->next->val == head->val){ ListNode* tmp = head; head = head->next; delete tmp; } return head; } }; |