LeetCode 92. Reverse Linked List II

Description:

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.

Submission Details

44 / 44 test cases passed.

Status: Accepted

Runtime: 3 ms

/**
* 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) {
ListNode* h = head;
ListNode* t = head;
ListNode* left;
ListNode* right;
for (int i = 1; i < m; i++) {
left = h; //翻转序列的前一个
h = h->next; //定位到翻转的第一个
}
for (int i = 1; i < n; i++) {
t = t->next; //定位到翻转的最后一个
}
right = t->next; //翻转序列的下一个
ListNode* pre = right;
ListNode* cur = h;
while (cur != right) {
ListNode* tmp = cur->next;
cur->next = pre;
pre = cur;
cur = tmp;
}
h->next = right;
if (m == 1)
return pre;
left->next = pre;
return head;
}
};