題目：
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
解法1：
迭代法
相比於83題將重複的元素只留一個，該題將重複的元素全部刪去
所以head節點也有可能會被刪去，所以要增加head節點前增加一個dummy節點
設置兩個指針pre和cur，初始指向dummy和head節點
遍歷鏈表，分兩種情況，一種是未遇到重複節點，這種情況，每次pre節點被賦爲當前節點cur，當前節點cur繼續向後移動一位
另一種遇到重複節點，這樣的話，一直遍歷重複的節點知道最後一個重複的節點，最後通過改變pre指針的位置來忽略這些相同的節點
記得在while循環中加上 head->next不爲空的判斷條件
c++：
/**
* 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* dummy = new ListNode(-1);
dummy->next = head;
ListNode* pre = dummy;
ListNode* cur = head;
while(cur && cur->next){
bool equal = false;
while(cur->next && cur->val == cur->next->val){
cur = cur->next;
equal = true;
}
if(equal){
pre->next = cur->next;
} else {
pre = cur;
}
cur = cur->next;
}
return dummy->next;
}
};
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 dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
ListNode cur = head;
while(cur!=null && cur.next!=null){
boolean equal = false;
while(cur.next!=null && cur.val == cur.next.val){
cur = cur.next;
equal = true;
}
if(equal){
pre.next = cur.next;
} else {
pre = cur;
}
cur = cur.next;
}
return dummy.next;
}
}
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
"""
dummy = ListNode(-1)
dummy.next = head
pre = dummy
cur = head
while cur and cur.next:
equal = False
while cur.next and cur.val == cur.next.val:
cur = cur.next
equal = True
if equal:
pre.next = cur.next
else:
pre = cur
cur = cur.next
return dummy.next
解法2：
遞歸法
相比於83題將重複的元素只留一個，該題將重複的元素全部刪去，所以相比於83題的if條件增加了while循環
83題：如果兩個節點相同，保留第二個節點
該題：如果兩個節點相同，繼續循環向後移動直到到達最後一樣相同的節點，然後捨棄這個節點從它的下一個節點繼續遞歸
記得在while循環中加上 head->next不爲空的判斷條件
c++：
/**
* 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){
while(head->next && head->val == head->next->val){
head = head->next;
}
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){
while(head.next != null && head.val == head.next.val){
head = head.next;
}
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:
while head.next and head.val == head.next.val:
head = head.next
return self.deleteDuplicates(head.next)
else:
head.next = self.deleteDuplicates(head.next)
return head