Remove Duplicates from Sorted List I and II

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题目1：

Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

代码1：

/**
 * 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 == nullptr)
            return nullptr;
        ListNode* cur = head;
        while(cur != nullptr)
        {
            if(cur->next == nullptr)
                break;
            if(cur->val == cur->next->val)
                cur->next = cur->next->next;
            else
                cur = cur->next;
        }
        return head;
    }
};

代码2：

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == nullptr)
            return nullptr;
        ListNode* pre = head;
        ListNode* cur = head;
        while(cur != nullptr)
        {
            while(cur != nullptr && pre->val == cur->val)
                cur = cur->next;
            pre->next = cur;
            pre = cur;
            if(cur != nullptr)
                cur = cur->next;
        }
        return head;
    }
};

---

题目2：

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

Input: 1->1->1->2->3
Output: 2->3

代码：

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == nullptr)
            return nullptr;
        ListNode* newHead = new ListNode(-1);
        newHead->next = head;
        ListNode* pre = newHead;
        ListNode* cur = head;
        while(cur != nullptr)
        {
            while(cur->next != nullptr && cur->val == cur->next->val)
                cur = cur->next;

            if(pre->next == cur)
                pre = pre->next;
            else
                pre->next = cur->next;

            cur = cur->next;
        }
        return newHead->next;
    }
};