[leetcode] 【链表】 82. Remove Duplicates from Sorted List II
2016-05-30 17:28
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Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given
return
Given
这一题的区别是这里要把重复的全部删除,比如
Given
return
Given
For example,
Given
1->2->3->3->4->4->5,
return
1->2->5.
Given
1->1->1->2->3, return
2->3.
题意
把链表中含有重复值元素的节点删掉,和Remove Duplicates from Sorted List这一题的区别是这里要把重复的全部删除,比如
Given
1->2->3->3->4->4->5,
return
1->2->5.
Given
1->1->1->2->3, return
2->3.
题解
两个指针,一个指向临时头,一个指向当前节点,如果没有重复元素,那么临时头和当前结点往后移动,如果有重复,把重复的都删掉,临时头的和被删掉的部分的下一个节点连接。语言描述不是很清晰,直接看代码会好点:/** * 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) return head; ListNode newhead(-1); newhead.next=head; ListNode *prev=&newhead; ListNode *cur=head; while(cur) { bool isDuplicate=false; while(cur->next!=NULL&&cur->val==cur->next->val) { isDuplicate=true; ListNode *temp=cur; cur=cur->next; delete temp; } if(isDuplicate) { ListNode *temp=cur; cur=cur->next; prev->next=cur; delete temp; } else { prev=prev->next; cur=cur->next; } } return newhead.next; } };
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