删除链表中等于给定值 val 的所有节点。

示例:

输入: 1->2->6->3->4->5->6, val = 6
输出: 1->2->3->4->5

Solution

• 借助虚拟头结点

// @lc code=start
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if(!head) return head;
        ListNode* dummyHead = new ListNode(-1, head);
        ListNode* cur = dummyHead;
        while(cur->next){
            if(cur->next->val==val){
                ListNode* delNode = cur->next;
                cur->next = delNode->next;
                delete delNode;
            }else{
                cur = cur->next;
            }
        }
        return dummyHead->next;
    }
};
// @lc code=end

• 递归法

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

java

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) {
            return head;
        }
        ListNode dummyHead = new ListNode(-1, head);
        ListNode cur = dummyHead;
        while (cur.next != null) {
            if (cur.next.val == val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return dummyHead.next;
    }
}

迭代法

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) {
            return head;
        }
        head.next = removeElements(head.next, val);
        return head.val == val ? head.next : head;
    }
}