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【LeetCode】Remove Duplicates from Sorted List II
Remove Duplicates from Sorted List II
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.
把非重复元素放置到新的链表中。
新链表设置表头newhead，以免纠结重复的元素是否为头结点的边界情况，反正返回的是newhead->next。
cur指针在原链表中遍历，如果与新链表尾newtail不同，则新链表新增元素；若相同，把newtail一并删去。
代码简单不多解释了。
/** * 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; else {//at least two nodes ListNode* cur = head->next; ListNode* newhead = new ListNode(-1); ListNode* newtail = new ListNode(head->val); newhead->next = newtail; ListNode* pretail = newhead; while(cur != NULL) { if(cur->val != newtail->val) { newtail->next = new ListNode(cur->val); cur = cur->next; pretail = newtail; newtail = newtail->next; } else { while(cur != NULL && cur->val == newtail->val) cur = cur->next; //cur == NULL or cur diffs newtail if(cur == NULL) { pretail->next = NULL; //delete newtail return newhead->next; } else { newtail = new ListNode(cur->val); pretail->next = newtail; cur = cur->next; } } } return newhead->next; } }};
有人指出我的算法创建了太多新节点，空间复杂度较高。
其实很好改，把原链表上的节点“拆”下来装到新链表尾部就行了。
/** * 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; else {//at least two nodes ListNode* cur = head->next; ListNode* newhead = new ListNode(-1); ListNode* newtail = new ListNode(head->val); newhead->next = newtail; ListNode* pretail = newhead; while(cur != NULL) { if(cur->val != newtail->val) { newtail->next = cur; cur = cur->next; pretail = newtail; newtail = newtail->next; newtail->next = NULL; //cut off } else { while(cur != NULL && cur->val == newtail->val) cur = cur->next; //cur == NULL or cur diffs newtail if(cur == NULL) { pretail->next = NULL; //delete newtail return newhead->next; } else { newtail = cur; pretail->next = newtail; cur = cur->next; newtail->next = NULL; //cut off } } } return newhead->next; } }};
【LeetCode】Remove Duplicates from Sorted List II
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