Leetcode刷题-203移除链表元素

#include
using namespace std;

//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) {
		ListNode *dummy = new ListNode(0);
		dummy->next = head;
		ListNode *pre = dummy;

		//方案一
		while (pre->next != NULL){
			if (pre->next->val == val) {
				pre->next = pre->next->next;
			}
			else{
				pre = pre->next;
			}
		}
		return dummy->next;

		//方案二（此方案存在bug,若最后一个值为val，无法删除）
		//while (head->next != NULL) {
		//	if (head->val == val) {
		//		pre->next = head->next;
		//		head = head->next;
		//	}
		//	else {
		//		pre = head;
		//		head = head->next;
		//	}
		//}
	}
};

ListNode* NUM(char num[], int size) {
	ListNode *L1 = new ListNode(NULL);
	ListNode *pre = L1;
	for (int i = 0; i < size; i++)
	{
		ListNode *p = new ListNode(NULL);
		p->val = num[i];

		pre->next = p;
		pre = p;
		p->next = NULL;
	}
	
	return L1->next;
}

void print(ListNode *L) {

	for (int i = 0; i < 5; i++)
	{
		cout << L->val << endl;
		L = L->next;
	}
}

void test01() {
	char num[] = { 1,2,6,3,4,5,6 };
	int m_size = sizeof(num);
	ListNode *L1 = NUM(num, m_size);

	Solution S;
	ListNode *L2 = S.removeElements(L1, 6);
	print(L2);
}

int main()
{
	test01();
	system("pause");
	return 0;
}