[LeetCode-82] Remove Duplicates from Sorted List II

来源：互联网发布：刘慈欣光荣与梦想知乎编辑：程序博客网时间：2024/05/29 09:13

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.

方法一

分析：新建一个链表，用新的链表，来接收剔除相同数字后的链表，相当于把剔除后的链表插入到新的链表中。（不符合题目要求）

struct ListNode* deleteDuplicates_newList(struct ListNode* head) {    if(!head) {return NULL;}struct ListNode* headTemp = head;struct ListNode* preListNodeTemp = NULL;struct ListNode* rListNodeHead = NULL;struct ListNode* rListNodepre = NULL;struct ListNode* rListNodecur = NULL;while(headTemp) {preListNodeTemp = headTemp;while(headTemp->next&&headTemp->val==headTemp->next->val) {headTemp = headTemp->next;}if(preListNodeTemp != headTemp&&headTemp->val == preListNodeTemp->val) {headTemp = headTemp->next;}if(preListNodeTemp == headTemp) {/*头结点*/if(rListNodeHead==NULL) {rListNodeHead = (struct ListNode*)malloc(sizeof(struct ListNode));rListNodepre = rListNodeHead;/*save the first linknode*/rListNodeHead->next  = NULL;/*set the linknode end*/rListNodeHead->val = headTemp->val;/*set the value*/}else {rListNodecur = (struct ListNode*)malloc(sizeof(struct ListNode));rListNodepre->next = rListNodecur;/*insert the linknode*/rListNodepre = rListNodepre->next;/*save the linknode*/rListNodepre->next = NULL;/*set the linknode end*/rListNodecur->val = headTemp->val;/*set the value*/}headTemp = headTemp->next;}}return rListNodeHead;}

方法二

题意要把链表中有重复的元素全部去除，只留下没有重复过的元素。
依旧是链表的操作，由于头节点也有可能删除，因此先建一个newListNodeHead节点头节点，方便原节点的删除。

struct ListNode* deleteDuplicates(struct ListNode* head) {    if(!head) {return NULL;}struct ListNode* headTemp = head;struct ListNode* freeListNodeTemp = NULL;struct ListNode* preListNodeTemp = NULL;/*1.creat a new Head*/struct ListNode* newListNodeHead = (struct ListNode*)malloc(sizeof(struct ListNode));struct ListNode* newListNodeHeadTemp = NULL;newListNodeHead->next = NULL;int valuePre = 0;/*2.Find the first LinkNode*/while(headTemp) {preListNodeTemp = headTemp;valuePre = preListNodeTemp->val;while(headTemp->next&&headTemp->val==headTemp->next->val) {freeListNodeTemp = headTemp;/*save the linknode to free*/headTemp = headTemp->next;/*move to next linknode*/free(freeListNodeTemp);/*free the linknode*/}if(preListNodeTemp != headTemp&&headTemp->val == valuePre) {freeListNodeTemp = headTemp;/*save the linknode to free*/headTemp = headTemp->next;/*move to next linknode*/free(freeListNodeTemp);/*free the linknode*/}if(preListNodeTemp == headTemp) {if(newListNodeHead->next==NULL) {newListNodeHeadTemp = headTemp;newListNodeHead->next = newListNodeHeadTemp;}else {newListNodeHeadTemp->next = headTemp;newListNodeHeadTemp = newListNodeHeadTemp->next;}headTemp = headTemp->next;}}if(newListNodeHeadTemp!=NULL) {newListNodeHeadTemp->next = NULL;}head = newListNodeHead->next;free(newListNodeHead);return head;}

再向前移动pre和now指针。

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