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.

/**
 * Definition for singly-linked list.
 * struct ListNode 
   {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) 
{
    if(NULL == head)
        return NULL;
    
    struct ListNode * p = head;
    while(p->next)
    {
        if(p->val != p->next->val)
        {
            p = p->next;
        }
        else
        {
            struct ListNode * tmp = p->next;
            p->next = p->next->next;
            /*free tmp,if node is malloc*/
            free(tmp);   
        }
            
    }
    
    return head;
    
}

经验教训

删除结点后，是否需要free应根据链表创建的方式来对待，比如struct ListNode a,b,c,d,e;然后各自赋值，串起来，就无需free,只有通过malloc申请的内存才需要free.

2021.05.23 C++实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* p = head;
        while(p&&p->next)
        {
            ListNode* next = p->next;
            if(p->val == next->val)
            {
                p->next = next->next;
            }else{
                p = p->next;
            }
        }
        return head;
    }
};

更简洁的实现：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* p = head;
        while(p)
        {
            while(p->next && p->val == p->next->val)
            {
                p->next = p->next->next;
            }
            p = p->next;
        }
        return head;
    }
};