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
为了最后能return修改后的链表，所以我们创建了一个总头dummy，来指向原来链表的head，然后再创造俩个指针，一前一后用来比较

/** * 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，不用记录Head
新建一个指针来代替head移动，然后释放删除节点。最后再比较第一个节点，然后正好返回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;    }};

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思路3
递归撒

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;}};