题目描述
在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。 例如，链表1->2->3->3->4->4->5 处理后为 1->2->5
思路：直接处理
1 /* 2 struct ListNode { 3 int val; 4 struct ListNode *next; 5 ListNode(int x) : 6 val(x), next(NULL) { 7 } 8 }; 9 */ 10 class Solution { 11 public: 12 ListNode* deleteDuplication(ListNode* pHead) 13 { 14 ListNode *res = new ListNode(0), *tmp = res; 15 ListNode *cur = pHead, *last; 16 bool flag = false; //判断一个元素是否重复 17 //从表头开始 18 while (cur != NULL) { 19 last = cur->next; 20 flag = false; 21 while ((last != NULL) && (cur->val == last->val)) { 22 last = last->next; 23 flag = true; 24 } 25 //如果有重复，那么cur指向下一个第一次不重复的节点 26 if (flag) { 27 cur = last; 28 } else { 29 tmp->next = cur; 30 tmp = cur; 31 //tmp->next = NULL; 千万注意不要在这里断掉，cur->next就会被置为NULL 32 cur = cur->next; 33 tmp->next = NULL; 34 } 35 } 36 return res->next; 37 } 38 };
以上代码内存未回收，更新版本如下：
1 /* 2 struct ListNode { 3 int val; 4 struct ListNode *next; 5 ListNode(int x) : 6 val(x), next(NULL) { 7 } 8 }; 9 */ 10 class Solution { 11 public: 12 13 ListNode* deleteDuplication(ListNode* pHead) 14 { 15 ListNode *res = new ListNode(0), *tmp = res; //res为表头，为方便链表一个节点重复的情况 16 ListNode *cur = pHead, *last = nullptr, *pNode; //pNode指向要回收的节点 17 bool flag = false; //判断某一个节点是否重复 18 while (cur != nullptr) { 19 flag = false; 20 last = cur->next; 21 while (last != nullptr && cur->val == last->val) { 22 pNode = last; 23 last = last->next; 24 25 delete pNode; 26 pNode = nullptr; 27 28 flag = true; 29 } 30 //如果有重复，那么cur指向下一个第一次不重复的节点 31 if (flag) { 32 pNode = cur; 33 cur = last; 34 35 delete pNode; 36 pNode = nullptr; 37 38 } else { 39 tmp->next = cur; 40 tmp = cur; 41 //tmp->next = NULL; 千万注意不要在这里断掉，cur->next就会被置为NULL 42 cur = cur->next; 43 tmp->next = nullptr; 44 } 45 } 46 return res->next; 47 } 48 }；
递归解法
1 class Solution { 2 public: 3 ListNode* deleteDuplication(ListNode* pHead) 4 { 5 if (pHead == NULL) 6 return NULL; 7 if (pHead != NULL && pHead->next == NULL) 8 return pHead; 9 ListNode *cur = pHead->next; 10 if (cur->val == pHead->val) { 11 cur = cur->next; 12 while (cur && cur->val == pHead->val) { 13 cur = cur->next; 14 } 15 return deleteDuplication(cur); 16 } else { 17 pHead->next = deleteDuplication(cur); 18 return pHead; 19 } 20 } 21 };
更新版本如下：
1 /* 2 struct ListNode { 3 int val; 4 struct ListNode *next; 5 ListNode(int x) : 6 val(x), next(NULL) { 7 } 8 }; 9 */ 10 class Solution { 11 public: 12 ListNode* deleteDuplication(ListNode* pHead) 13 { 14 if (pHead == nullptr) { 15 return nullptr; 16 } 17 if (pHead != nullptr && pHead->next == nullptr) { 18 return pHead; 19 } 20 ListNode *cur = pHead->next; 21 if (cur->val == pHead->val) { 22 ListNode *tmp = cur; 23 cur = cur->next; 24 delete tmp; 25 tmp = nullptr; 26 while (cur != nullptr && cur->val == pHead->val) { 27 ListNode *tmp = cur; 28 cur = cur->next; 29 delete tmp; 30 tmp = nullptr; 31 } 32 return deleteDuplication(cur); 33 } else { 34 pHead->next = deleteDuplication(cur); 35 return pHead; 36 } 37 } 38 };