20190601-剑指offer18.2-AcWing-29. 删除链表中重复的节点

剑指offer18.2-AcWing-29. 删除链表中重复的节点

在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留。
样例1
输入：1->2->3->3->4->4->5
输出：1->2->5
样例2
输入：1->1->1->2->3
输出：2->3

思路：
首先判断节点是否为空，若是空，则直接return
初始两个变量，一个指向当前节点nowNode，一个指向当前节点的前一个结点preNode。
不管循环遍历当且结点：
  若当前结点与下一个结点重复，则删除，直到重复结点全部删除
  比较巧妙的设计为用preNode是否为空，决定head结点。
  连接preNode与nowNode
C++ code：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplication(ListNode* head) {
        if(head == NULL){
            return head;
        }
        ListNode* preNode = NULL;
        ListNode* nowNode = head;
        while(nowNode != NULL){
            //ListNode* nextNode = nowNode->next;
            bool is_del = 0;
            if(nowNode->next != NULL && nowNode->val == nowNode->next->val){
                is_del = 1;
            }
            if(!is_del){
                preNode = nowNode;
                nowNode = nowNode->next;
            }else{
                int value = nowNode->val;
                while(nowNode != NULL && nowNode->val == value){
                    ListNode* temp = nowNode;
                    nowNode = nowNode->next;
                    delete temp;
                }
                
                if(preNode == NULL){
                    head = nowNode;
                }else{
                    preNode->next = nowNode;
                }
            }
        }
        return head;
    }
};

python code：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplication(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None:
            return head
        
        preNode = None
        nowNode = head
        
        while nowNode != None:
            
            is_del = 0
            if nowNode.next != None and nowNode.val == nowNode.next.val:
                is_del = 1
                
            if not is_del:
                preNode = nowNode
                nowNode = nowNode.next
            else:
                value = nowNode.val
                while nowNode != None and nowNode.val == value:
                    temp = nowNode
                    nowNode = nowNode.next
                    del temp
                
                if preNode == None:
                    head = nowNode
                else:
                    preNode.next = nowNode
                    
        return head