31：Remove Duplicates from Sorted List II

本题解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解

题目：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.

解题代码版本一：

//迭代版本//时间复杂度 O(n)，空间复杂度O(1)class Solution {public:        ListNode* deleteDuplicates(ListNode* head) {                if (head == nullptr) return head;                ListNode dummy(-1); // 头结点                dummy.next = head;                ListNode *prev = &dummy, *cur = head;                while (cur != nullptr) {                        bool duplicated = false;                        while (cur -> next != nullptr && cur -> val == cur -> next -> val) {                                duplicated = true;                                ListNode *temp = cur;                                cur = cur -> next;                                delete tmp;                        }                        if (duplicated) { //删除重复的最后一个元素                                ListNode* temp = cur;                                cur = cur -> next;                                delete temp;                                prev -> next = cur;                        }                        else {                                prev -> next = cur;                                prev = prev -> next;                                cur = cur -> next;                        }                }                return dummy.next;        }};

解题代码版本二：

//递归版本//时间复杂度O(n)，空间复杂度O(1)class Solution {public:        ListNode* deleteDuplicates(ListNode* head) {                if (head == nullptr || head -> next == nullptr) return head;                ListNode* p = head -> next;                if (head -> val == p -> val) {                        while (p && head -> val == p -> val) {                                ListNode* tmp = p;                                p = p -> next;                                delete tmp;                        }                        delete head;                        return deleteDuplicate(p);                }                else {                        head -> next = deleteDuplicates(head -> next);                        return head;                }        }};