203. Remove Linked List Elements【easy】

203. Remove Linked List Elements【easy】

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

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

解法一：

 /**
  * 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;
         }

         ListNode * dummy = new ListNode(INT_MIN);
         dummy->next = head;
         head = dummy;

         while (head != NULL && head->next != NULL) {
             if (head->next->val == val) {
                 while (head->next != NULL && head->next->val == val) {
                     ListNode * temp = head->next;
                     free(temp);
                     head->next = head->next->next;
                 }
             }
             else {
                 head = head->next;
             }
         }

         return dummy->next;
     }
 };

里面的while可以不用，因为这个题和（82. Remove Duplicates from Sorted List II）不一样，那个题你根本不知道val是什么，所以只用一个if是不行的。但是这个题你知道val是什么，所以可以省略里面的while。

解法二：

 /**
  * 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;
         }

         ListNode * dummy = new ListNode(INT_MIN);
         dummy->next = head;
         head = dummy;

         while (head != NULL && head->next != NULL) {
             if (head->next->val == val) {
                 ListNode * temp = head->next;
                 free(temp);
                 head->next = head->next->next;
             }
             else {
                 head = head->next;
             }
         }

         return dummy->next;
     }
 };

精简写法

解法三：

 public ListNode removeElements(ListNode head, int val) {
         if (head == null) return null;
         head.next = removeElements(head.next, val);
         return head.val == val ? head.next : head;
 }

递归，参考了@renzid 的代码