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->6->NULL, m = 2 and n = 5,

return 1->5->4->3->2->6->NULL.

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.

解题思路：

（1）找到结点m的前驱结点记为preNode_m，若结点m为头结点则preNode_m=NULL；找到结点m，记为pNode_m；

（2）将结点m至n之间的链表反转

（3）找到结点n及其后继结点，分别记为pNode_n和nextNode_n

（4）改变指针的指向：结点m的后继结点指向结点n的后继结点；

（5）若结点m的前驱结点为NULL，则将结点n定义为新的头结点；否则，将结点m的前驱结点指向结点n

struct ListNode* reverseBetween(struct ListNode* head, int m, int n) {

struct ListNode* pNode = head;
struct ListNode* preNode_m = NULL;//结点m的前驱结点
for (int i = 1; i < m; i++)
{
preNode_m = pNode;
pNode = pNode->next;
}
struct ListNode* pNode_m = pNode;//结点m

struct ListNode* preNode = NULL;
for (int i = m; i <= n; i++)
{
struct ListNode* pNext = pNode->next;
pNode->next = preNode;

preNode = pNode;
pNode = pNext;
}
struct ListNode* pNode_n = preNode;//结点n
struct ListNode* nextNode_n = pNode;//结点n的后继结点

pNode_m->next = nextNode_n;//结点m的后继结点指向结点n的后继结点
if (preNode_m == NULL)
{
return pNode_n;
}
else
{
preNode_m->next = pNode_n;//结点m的前驱结点指向结点n
return head;
}

}