題目

給定連結串列的頭節點head，實現刪除連結串列中間節點的函式和刪除位於a/b處節點的函式。 當有偶數個節點時，刪除兩個中間節點的前一個，有奇數個節點時，刪除中間節點。
刪除a/b處的節點：連結串列1->2->3->4->5，假設a/b的值為r，如果r = 0，不刪除任何節點。如果r在區間（0，1/5]上，刪除節點1；依次類推；如果r>1，則不刪除任何節點。
具體題目可參考原書

程式碼

#include<iostream>
using namespace std;

struct node {
	int value;
	node* next;
 

	node(int val)
	{
		value = val;
	}
};

node* removeMidNode(node* head)
{
	if (head == NULL || head->next == NULL)
		return head;
	if (head->next->next == NULL)
		return head->next;
	node* prev = head;
	node* cur = head->next->next;
	while (cur->next != NULL && cur->
 
next->next != NULL)
	{
		prev = prev->next;
		cur = cur->next->next;
	}
	prev->next = prev->next->next;
	return head;
}

node* removeByRation(node* head, int a, int b)
{
	if (a < 1 || a > b)
		return head;
	int n = 0;
	node* cur = head;
	while (cur != NULL)
	{
		n++;
		cur =
 
 cur->next;
	}
	if (n == 1)
		head = head->next;
	if (n > 1)
	{
		n = ceil((double)(a * n) / (double)b);
		cur = head;
		while (--n != 1)
			cur = cur->next;
		cur->next = cur->next->next;
	}
	return head;
}

node* createList(int* in, int len)
{
	node* head = new node(in[0]);
	node* p = head;
	for (int i = 1; i < len; i++)
	{
		node* pNode = new node(in[i]);
		p->next = pNode;
		p = p->next;
	}
	p->next = NULL;
	return head;
}

void printList(node* pHead)
{
	cout << "Node of List:" << endl;
	if (pHead == NULL)
		cout << "NULL" << endl;
	while (pHead != NULL)
	{
		cout << pHead->value << " ";
		pHead = pHead->next;
	}
	cout << endl;
}

int main()
{
	int input1[] = { 1, 2, 3, 4, 5 };
	int input2[] = { 1, 2, 3, 4 };
	int input3[] = { 1, 2, 3, 4, 5, 6, 7 };
	int a = 5, b = 6;
	node* pHead1 = createList(input1, 5);
	node* pHead2 = createList(input2, 4);
	node* pHead3 = createList(input3, 7);
	node* p1 = removeMidNode(pHead1);
	node* p2 = removeMidNode(pHead2);
	node* p3 = removeByRation(pHead3, a, b);
	printList(p1);
	printList(p2);
	printList(p3);
	getchar();
	return 0;
}