本题目比較直接，一次遍历遇到匹配的元素直接删除（通过n.next = n.next.next）就能够了，仅仅是须要考虑到：

1.首节点的情况

2.末节点的情况

下面为实现：

public ListNode RemoveElements(ListNode head, int val) {        // null list	if(head == null){		return null;	}	// constains only one node	if(head.next == null && head.val == val){		return null;	}		//remove first nodes	while(head.val == val){	    if(head.next == null){	        break;	    }		head = head.next;	}	var tmp = head;		// nodes in between	while(head.next != null){		if(head.next.val == val){			head.next = head.next.next;		}		else{			head = head.next;		}		if(head.next == null){			break;		}	}	// last node	if(head.val == val){		return null;	}		// restore head node	head = tmp;		return head;	    }