LeetCode之“链表”：Remove Duplicates from Sorted List && Remove Duplicates from Sorted List II

　 1. Remove Duplicates from Sorted List

　　题目链接

　　题目要求：

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

　　For example,
　　Given 1->1->2, return 1->2.
　　Given 1->1->2->3->3, return 1->2->3.

　　这道题不难。具体程序如下：

/**
 * 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 || !head->next)
            return head;
        
        int preVal = head->val;
        ListNode *preNode = head;
        ListNode *start = head->next;
        while(start)
        {
            while(start && start->val == preNode->val)
            {
                ListNode *next = start->next;
                preNode->next = next;
                delete start;
                start = next;
            }
            
            if(!start)
                break;
                
            preNode = start;
            start = start->next;
        }
        
        return head;
    }
};

　 2. Remove Duplicates from Sorted List II  

　　题目链接

　　题目要求：

　　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.

　　该题可以通过添加dummy节点以方便编程。具体程序如下：

/**
 * 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 || !head->next)
            return head;
        
        int preVal = head->val;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *prepreNode = dummy, *preNode = head;
        ListNode *start = head->next;
        while(start)
        {
            bool flag = true;
            while(start && start->val == preNode->val)
            {
                flag = false;
                ListNode *next = start->next;
                preNode->next = next;
                delete start;
                start = next;
            }
            
            if(flag)
            {
                prepreNode = preNode;
                preNode = preNode->next;
            }
            else
            {
                prepreNode->next = preNode->next;
                delete preNode;
                preNode = prepreNode->next;
            }
            
            if(!preNode || !preNode->next)
                break;
            start = preNode->next;
        }
        
        head = dummy->next;
        delete dummy;
        dummy = nullptr;
        
        return head;
    }
};