LEETCODE: 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.
今天写代码的状态真是不好！这道题居然写了很久。不过打完收工，睡觉去了。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
if(m == n) return head;

// If m == 1, means that a new head should be return;
ListNode *newhead = NULL;
// Find the last one before m
ListNode *prereverse = NULL;
for(int ii = 0; ii < m - 1; ii ++) {
if(ii == 0) {
prereverse = head;
}
else {
prereverse = prereverse->next;
}
}

// Reverse nodes from m to n.
ListNode *newshortend = m == 1 ? head : prereverse->next;
ListNode *start = m == 1 ? head : prereverse->next;
ListNode *end = NULL;
ListNode *current = start->next;
for(int ii = m; ii < n; ii ++) {
end = current->next;
ListNode *temp = current->next;
current->next = start;
start = current;
current = temp;
}

if(m == 1) {
newhead = start;
newshortend->next = current;
return newhead;
}
else {
prereverse->next = start;
newshortend->next = end;
return head;
}
}
};