Leetcode 92. Reverse Linked List II

原题：

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.

解决方法：

翻转链表的变种，需要翻转n-m+1个节点，也可以看成，从m开始,需要将后面的n-m个节点提到n-1的节点后面。

代码：

ListNode* reverseBetween(ListNode* head, int m, int n) {
if (!head || m==n)
return head;

ListNode dummy(INT_MIN), *prev=&dummy;
prev->next = head;

int index = 1;
while(head){
if (index >= m && index < n){
ListNode* cur = head;
ListNode* next = head->next;

cur->next = next->next;
next->next = prev->next;
prev->next = next;
}else if (index >=n){
break;
}else{
prev = prev->next;
head = head->next;
}

++index;
}

return dummy.next;
}