203. Remove Linked List Elements

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

Solution 1: add a sentinel to make the two conditions removing at the beginning and removing in the middle to be one condition. 

/**
 * 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* sentinel=new ListNode(-1);
        ListNode* pre=sentinel, *cur=head;
        sentinel->next=head;
        while(cur){
            if (cur->val==val){
                pre->next=cur->next;
                cur=pre->next;
            }
            else {
                cur=cur->next;
                pre=pre->next;
            }
        }
        return sentinel->next;
    }
};

Solution 2: only use one pointer. line 10 free the memory of the deleted node to avoid the memory leak.

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* sentinel=new ListNode(-1), *pre=sentinel;
        sentinel->next=head;
        while(pre->next){
            if (pre->next->val==val){
                ListNode* delptr=pre->next;
                pre->next=pre->next->next;
                delete delptr;
            }
            else pre=pre->next;
        }
        return sentinel->next;
    }
};

 Solution 3: use recursion

class Solution {
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;
    }
};

consider freeing the deleted node memory

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if (head==NULL) return NULL;
        head->next = removeElements(head->next, val);
        if (head->val==val){
            ListNode* next=head->next;
            delete head;
            return next;
        }
        else return head;
    }
};