Reverse Linked List II--チェーンテーブルとポインタを鍛える問題

タイトルリンク:https://oj.leetcode.com/problems/reverse-linked-list-ii/

#include 
using namespace std;

struct ListNode {
	int val;
	ListNode *next;
	ListNode(): val(0), next(NULL) {}
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	ListNode *reverseBetween(ListNode *head, int m, int n) {
		if(head == NULL || m == n) return head;
		int i;
		ListNode *first = NULL;
		ListNode *temp  = head;
		for(i = 0; i < m - 1; i++)
		{
			first = temp;
			temp  = temp -> next;
		}
		ListNode *end      = temp;
		ListNode *temp_pre = temp;
		temp               = temp -> next;
		for(i = m + 1; i <= n; i++)                  //    ：     next         ，  temp_pre temp    
		{
			ListNode *temp_next = temp -> next;
			temp -> next        = temp_pre;
			temp_pre            = temp;
			temp                = temp_next;
		}
		end -> next = temp;                         //     m     n+1     
		if(first) {                                 //           
			first -> next = temp_pre;
		}
		else {
			head = temp_pre;
		}
		return head;
	}
};

int main() {
	int len,m,n,t,i,j;
	cin>>len;
	ListNode *node = new ListNode[len+1], *head;
	head = node;
	for(i = 0; i < len;i++) {
		cin>>t;
		j = i;
		node[i].val  = t;
		node[i].next = &node[++j];
	}
	cin>>m>>n;
	Solution a;
	ListNode *res = a.reverseBetween(head,m,n);
	for(i = 0; i < len; i++) {
		if(i != len - 1) {
			cout< val<";
			res = res -> next;
		}
		else {
			cout< val<