Leetcode Reverse Linked List II 反转特定区间的链表

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.
这样的题目关键是不要怕麻烦，一定要画多几个链表图，一步一步演算，这样思路就很清晰了。
通过头痛折磨的训练之后，这些题目不用调试，一次性AC。

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

class Solution {
public:
	ListNode *reverseBetween(ListNode *head, int m, int n) 
	{
		ListNode *cur = head;
		ListNode dummy(0);
		dummy.next = head;
		ListNode *pre = &dummy;

		for (int i = 1; i < m; i++)
		{
			pre = cur;
			cur = cur->next;
		}

		ListNode *post;
		ListNode *ppost;
		if (cur) ppost = cur;
		if (ppost) post = ppost->next;
		for (int j = m; j <= n; j++)
		{
			ppost->next = cur;
			cur = ppost;
			ppost = post;
			if (post) post = post->next;
		}
		pre->next->next = ppost;
		pre->next = cur;

		return dummy.next;
	}
};

上面是逆转法，下面是插入法 2014-1-10更新：

ListNode *reverseBetween(ListNode *head, int m, int n) 
	{
		if (!head || m==n) return head;
		ListNode dummy(-1);
		dummy.next = head;
		ListNode *pre = &dummy;

		for (int i = 1; i < m; i++)
		{
			pre = pre->next;
		}

		ListNode *insertPre = pre->next;
		for (int i = m; i < n; i++)
		{
			ListNode *cur = insertPre->next;
			insertPre->next = cur->next;
			cur->next = pre->next;
			pre->next = cur;
		}
		return dummy.next;
	}

//2014-2-14 update
	ListNode *reverseBetween(ListNode *head, int m, int n) 
	{
		ListNode dummy(0);
		ListNode *p = &dummy;
		dummy.next = head;

		for (int i = 1; i < m; i++) p = p->next;
		head = p->next;

		for ( ; m < n; m++)
		{
			ListNode *tmp = head->next;
			head->next = tmp->next;
			tmp->next = p->next;
			p->next = tmp;
		}
		return dummy.next;
	}