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

思路1
爲了最後能return修改後的鏈表，所以我們創建了一個總頭dummy，來指向原來鏈表的head，然後再創造倆個指針，一前一後用來比較

/**
 * 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(0);
        dummy->next = head;
        ListNode* p = dummy;
        ListNode* q = head;
        while(q!=NULL)
        {
            if(q->val == val)
                p->next = q->next;
            else
                p = p->next;

            q = q->next;
        }

        return dummy->next;

    }
};

思路2
和上面相反的方法，不用記錄Head
新建一個指針來代替head移動，然後釋放刪除節點。最後再比較第一個節點，然後正好返回Head

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if (head==NULL){
            return head;
        }
        ListNode* p=head;
        while(p->next!=NULL){
            if (p->next->val == val){
                auto freeNode = p->next;
                p->next = p->next->next;
                free(freeNode);
            }
            else{
                p=p->next;
            }
        }
        if (head->val==val)
            head=head->next;
        return head;
    }
};

---

思路3
遞歸撒

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