LeetCode | Remove Duplicates from Sorted List II（删除链表中重复结点2）

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

.

题目解析：

这道题看似简单，但却越到了很大的麻烦。时常少些必要的条件，经过修修补补，终于通过了……但不是一个好的方法。先贴上笨蛋方法……

利用flag来标记是否重复，以便后面进行修正。

class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
int flag = 0;

if(head == NULL || head->next == NULL)
return head;

ListNode *p,*q;
p = head;
q = p->next;

head = NULL;
while(q != NULL){
if(p->val == q->val){
flag = 1;
q = q->next;
continue;
}
if(flag == 1){
flag = 0;
p->val = q->val;
p->next = q->next;
if(head == NULL)
head = p;
q = q->next;
continue;
}
if(head == NULL)
head = p;
p->next = q;
p = q;
q = q->next;
}

if(head == NULL || (flag && head == p))
return NULL;
if(flag){
ListNode *h = head;
while(h->next != p)
h = h->next;
h->next = NULL;
}
return head;
}
};

方案二：

很好的方法，利用二级指针来修改。ppNext = &(head->next)。这样赋值以后，便不随head的改变而改变，其就是一个指针的地址。

class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
ListNode *root = NULL;
ListNode **ppNext = &root;
while(head) {
if(head->next == NULL || head->next->val != head->val) {
*ppNext = head;
ppNext = &(head->next);
head = head->next;
} else {
int val = head->val;
do {
head = head->next;
} while(head != NULL && head->val == val);
}
}
*ppNext = NULL;
return root;
}
};

或者：

class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(!head) return head;

int val = head->val;
ListNode ** pp = &head;
ListNode * cur = head;

if( cur->next == NULL || cur->next->val != cur->val  ) {
*pp = cur;
pp = &(cur->next);
}

while( (cur=cur->next) != NULL  ) {
if( cur->val != val) {
if( cur->next == NULL || cur->next->val != cur->val) {
*pp = cur;
pp = &(cur->next);
}
val = cur->val;
}
}

*pp = NULL;
return head;
}
};