【LeetCode】Reverse Linked List II

Reverse Linked List II

 Total Accepted: 15366 Total
Submissions: 59341My Submissions

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.

确定一个新链表的是首尾指针是一个非常不错的选择。

/**
* 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 (!head || !head->next) return head;
if (m < 1 || n <= 1 || (n - m) < 1) return head;
ListNode *tmp_head = NULL, *tmp_tail = NULL;
ListNode *save_tail = NULL;
ListNode *p = head;
int i = 1;
//找到m前一个结点
while (p && i < (m - 1)) {
p = p->next;
i++;
}
if (p == NULL || p->next == NULL) return head;
if (m > 1) {
save_tail = p;
p = p->next;
i += 1;
}
i += 1;
tmp_head = tmp_tail = p;
p = tmp_head->next;
while(p && i <= n) {
tmp_tail->next = p->next;
p->next = tmp_head;
tmp_head = p;
p = tmp_tail->next;
i++;
}
if (save_tail) {
save_tail->next = tmp_head;
} else {
head = tmp_head;
}
return head;
}
};