LeetCode - M - Reverse Linked List II

题意

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.

解法

记录翻转部分之前的一个节点pre，其他部分和单纯翻转一样

实现

/**
* 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 == NULL || m == n) return head;
ListNode* pre = NULL;
ListNode* cur = head;
int idx = 1;
while(idx < m && cur != NULL){
++idx;
pre = cur;
cur = cur->next;
}
ListNode* shead = cur;
while(idx < n && cur->next != NULL){
ListNode* next = cur->next;
cur->next = next->next;
next->next = shead;
shead = next;
++idx;
}
if(pre == NULL) head = shead;
else pre->next = shead;
return head;
}
};