给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。
示例 1:
输入: 1->2->3->3->4->4->5 输出: 1->2->5
示例 2:
输入: 1->1->1->2->3 输出: 2->3
1 #include "_000库函数.h" 2 3 4 struct ListNode { 5 int val; 6 ListNode *next; 7 ListNode(int x) : val(x), next(NULL) {} 8 }; 9 10 11 //又是无头结点的链表！！！ 12 //还有，哪个数字重复了，就将他全部删除！！ 13 class Solution { 14 public: 15 ListNode* deleteDuplicates(ListNode* head) { 16 //创建一个头结点 17 ListNode *p = new ListNode(0); 18 p->next = head; 19 head = p; 20 ListNode *q = p; 21 int flag = 0;//用来标记重复数字 22 while (q && p) { 23 q = q->next; 24 while (q && q->next && q->val == q->next->val) { 25 //一定的记得释放删除的数字 26 ListNode *ptr = q; 27 p->next = q->next; 28 delete(ptr); 29 ptr = NULL; 30 q = p->next; 31 flag = 1; 32 } 33 if (flag) {//删除最后一个重复了的数字 34 ListNode *ptr = q; 35 p->next = q->next; 36 delete(ptr); 37 ptr = NULL; 38 q = p; 39 flag = 0; 40 } 41 else 42 p = p->next; 43 } 44 return head->next; 45 } 46 }; 47 48 //不释放删除的节点，我不建议，但好多博客是这样写的 49 class Solution { 50 public: 51 ListNode* deleteDuplicates(ListNode* head) { 52 //创建一个头结点 53 ListNode *p = new ListNode(0); 54 p->next = head; 55 head = p; 56 ListNode *q = p; 57 while (p->next) { 58 q = p->next; 59 while (q->next && q->val == q->next->val) 60 q = q->next; 61 if (p->next != q) 62 p->next = q->next; 63 else 64 p = p->next; 65 } 66 return head->next; 67 } 68 }; 69 70 71 72 //使用递归，不使用while 73 74 class Solution { 75 public: 76 ListNode *deleteDuplicates(ListNode *head) { 77 if (!head) return head; 78 if (head->next && head->val == head->next->val) { 79 while (head->next && head->val == head->next->val) { 80 head = head->next; 81 } 82 return deleteDuplicates(head->next); 83 } 84 head->next = deleteDuplicates(head->next); 85 return head; 86 } 87 }; 88 void T082() { 89 ListNode *head = new ListNode(0); 90 ListNode *p = head; 91 vector<int>v = { 1,2,3,4,4, 5, 5}; 92 for (auto a : v) { 93 ListNode *q = new ListNode(0); 94 q->val = a; 95 p->next = q; 96 p = q; 97 } 98 p = head->next; 99 while (p) { 100 cout << p->val << "->"; 101 p = p->next; 102 } 103 cout << endl; 104 Solution s; 105 p = s.deleteDuplicates(head->next); 106 while (p) { 107 cout << p->val << "->"; 108 p = p->next; 109 } 110 cout << endl; 111 }