LeetCode 刷题记录 83. Remove Duplicates from Sorted List

题目：
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
解法1：
迭代法：
c++：
遍历链表，一次遍历两个数。如果两个数相同，更改指针的位置将第二个数逻辑上删去
如果不同，继续下一个数再次进行循环

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

java：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode cur = head;
        while(cur!=null && cur.next!=null){ 
        
            if(cur.val == cur.next.val){ 
                cur.next = cur.next.next; 
       
            }else{
                cur = cur.next;
            }
        }
        return head;
    }
}

python：

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

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        cur = head
        while cur and cur.next: 
        
            if cur.val == cur.next.val:
                cur.next = cur.next.next 
       
            else:
                cur = cur.next
            
        
        return head

解法2：递归：
递归基：
如果只有一个结点或者没有结点，则返回
判断当前结点和它的下一个结点的值，如果相等，则舍弃当前结点，继续递归它的下一个结点
如果不等，继续递归它的下一个结点，但是要通过指针操作将当前结点与下一个结点接上，并返回当前结点

/**
 * 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;
        if(head->val == head->next->val){
                return deleteDuplicates(head->next);
        } else {
                head->next = deleteDuplicates(head->next);
        }
       
        return head;
    }
};

java：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
         if(head == null || head.next == null){
            return head;
        }
    
        if(head.val == head.next.val){
        
            return deleteDuplicates(head.next);
        }else{
        
            head.next = deleteDuplicates(head.next);
            
        }
        return head;
    }
        
}

python：

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

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head  or not head.next:
            return head
        
    
        if head.val == head.next.val:
        
            return self.deleteDuplicates(head.next)
        else:
        
            head.next = self.deleteDuplicates(head.next)
            
        
        return head