[leetcode]Reverse Linked List II

新博文地址： [leetcode]Reverse Linked List II

http://oj.leetcode.com/problems/reverse-linked-list-ii/

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

 在剑指offer中看到过反转链表，里面讲的思想稍稍复杂一点，是在原链表中进行反转操作，但是我拿到这道题的第一印象就是用头插法生成新的链表，并将新建链表查到原来链表的相应位置中。思想很简单明了，但是代码略长。

public ListNode reverseBetween(ListNode head, int m, int n) {
    	if(m == n || head == null){
    		return head;
    	}
    	//------ 手动添加头结点，为了操作统一性
    	ListNode phead = new ListNode(0);
    	phead.next = head;
    	ListNode tem = phead;
    	
    	int count = 0;
    	//需要记录m ，n前后的两个节点，方便新链表的插入
    	ListNode newList = new ListNode(0);
    	ListNode mPreNode = null;
    	ListNode nPreNode = null;
    	ListNode tail = null;
    	
    	while(tem != null) {
    		if(count == m - 1){
    			if(m == 1){
					mPreNode = phead;
    			}else{
    				mPreNode = tem;
    			}
    		}else if (count == n + 1){
    			nPreNode = tem;
    		}else if(count <= n && count >= m){
                        //头插法建表
    			ListNode node = new ListNode(tem.val);
    			node.next = newList.next;
    			newList.next = node;
    			if(count == m){
    				tail = node;
    			}
    		}
    		count++;
    		tem = tem.next;
    	}
    	mPreNode.next = newList.next;
    	tail.next = nPreNode;
    	return phead.next;
    }