LeetCode OJ 之 Remove Linked List Elements（删除链表中的元素）

题目：

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

思路：

1、迭代

2、递归

代码1：

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val)
{
ListNode *dummy = new ListNode(-1);//虚拟头结点，因为有可能把原链表的头结点删掉
ListNode *p1 = dummy , *p2 = head;
while(p2)
{
if(p2->val == val)
{
ListNode *tmp = p2;
p2 = p2->next;
delete tmp;
}
else
{
p1->next = p2;
p1 = p1->next;
p2 = p2->next;
}
}
p1->next = NULL;//这一步很重要，切记
return dummy->next;
}
};

代码2：

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val)
{
if(head == NULL)
return head;
head->next = removeElements(head->next , val);
return head->val == val ? head->next : head;
}
};