LeetCode 203. 移除链表元素

题目描述：

解法一（自然想法）：

/**
 * 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* pre = NULL, *now = head, *res = head;
        while (now != NULL) {
            if (now->val == val) {          
                if (now == res) {                //过滤头节点
                    res = now->next;
                    now = res;

                }
                else {                            //过滤平常结点
                    pre->next = now->next;
                    now = now->next;
                }
            }
            else {
                pre = now;
                now = now->next;
            }
        }
        return res;
    }
};

解法二（删除头结点时另做考虑（由于头结点没有前一个结点））:

/**
 * 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) {
        while(head!=NULL&&head->val==val) head=head->next;        //滤除头节点
        if(head==NULL) return head;
        ListNode* now=head;
        while(now->next!=NULL){
            if(now->next->val==val){                              //滤除普通节点
                now->next=now->next->next;
            }
            else
                now=now->next;
        }
        return head;
    }
};

解法三（添加一个虚拟头结点，删除头结点就不用另做考虑 ）：

/**
 * 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* nHead=new ListNode(val-1);                //添加虚拟头结点
        nHead->next=head;
        ListNode* now=nHead;
        while(now->next!=NULL){
            if(now->next->val==val){
                now->next=now->next->next;
            }
            else
                now=now->next;
        }
        ListNode* res=nHead->next;
        delete nHead;
        return res;
    }
};

解法四（递归）：

/**
 * 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 NULL;
        head->next=removeElements(head->next,val);
        if(head->val==val){
            return head->next;
        }
        else return head;
    }
};