[LeetCode]Reverse Linked List II

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

class Solution {
//divide this problem into 3 parts
//1. previous part, do not need to reverse
//2. median part, need to reverse
//3. post part, do not need to reverse
//note:
//1. link the previous part to median part & link median part to post part
//2. get the new head
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ListNode* pre = NULL;
ListNode* median = NULL;
ListNode* post = NULL;
ListNode* p = head;
//1. previous part
for (int i = 1; i < m && p; ++i)
{
if(!pre)
pre = p;
else
{
pre->next = p;
pre = pre->next;
}
p = p->next;
}
//2. median part
ListNode* firstMedian = NULL;
for (int i = m; i <= n && p; ++i)
{
if(!median)
{
median = p;
p = p->next;
median->next = NULL;
firstMedian = median;
}
else
{
ListNode* tmp = p->next;
p->next = median;
median = p;
p = tmp;
}
}
if(pre) pre->next = median;
if(firstMedian) firstMedian->next = p;
//find the head
if(!pre)
head = median;
return head;
}
};

/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ListNode dummy(-1);
dummy.next = head;
ListNode* prev = &dummy;//node before reverse range
ListNode* cur = head;
ListNode* reverseTail;
ListNode* reverseHead;
int curIdx = 1;
while(cur != NULL)
{
if(curIdx == m) reverseHead = cur, reverseTail = cur;
else if(m < curIdx && curIdx <= n)
{
ListNode* tmp = cur->next;
prev->next = cur;
cur->next = reverseHead;
reverseHead = cur;
reverseTail->next = tmp;
cur = reverseTail;
}
else if(curIdx < m) prev = cur;
else break;//do not need to reverse anymore

cur = cur->next;
curIdx++;
}
return dummy.next;
}
};