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