[LeetCode] Reverse Linked List II 解题报告

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.

» Solve this problem

[解题思路]
分三步走
1. 找到m节点的前一个指针pre（加个safe guard可避免头指针的问题）
2. 从m节点开始往后reverse N个节点（双指针，cur，post）
3. 合并pre链表，cur链表及post链表。

这题难就难在繁琐上，要考虑各种边界条件，比如
{1,2,3}, 3,3
{1,2,3}, 1,1
{1,2,3}, 1,3
所以，code中需要添加一些边界检查条件。这题15分钟以内bug free我是做不到。

1:       ListNode *reverseBetween(ListNode *head, int m, int n) {   
2:            // Start typing your C/C++ solution below   
3:            // DO NOT write int main() function   
4:            int step = n-m;   
5:            ListNode* safeG = new ListNode(-1); //intro a safe guard to avoid handle head case  
6:            safeG->next = head;   
7:            head = safeG;   
8:            ListNode* pre = head;   
9:            while(m>1)   
10:            {   
11:                 pre=pre->next;   
12:                 m--;   
13:            }   
14:            ListNode* cur = pre->next, *post = cur->next;   
15:            if(step>=1)   
16:            {   
17:                 while(step>0 && post!=NULL)   
18:                 {   
19:                      ListNode* temp = post->next;   
20:                      post->next = cur;   
21:                      cur = post;   
22:                      post = temp;   
23:                      step--;   
24:                 }   
25:                 ListNode* temp = pre->next;   
26:                 pre->next = cur;   
27:                 temp->next = post;   
28:            }   
29:            safeG = head;   
30:            head = head->next;   
31:            delete safeG;   
32:            return head;     
33:       }   

Update: 3/19/2013. SafeG在这里没有意义。

Update 3/6/2014. 又看了一遍，SafeG还是有必要的，避免对于head的处理。