标签:c style class blog code java
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.
先找到需要翻转的起始节点,然后,翻转其后的n - m 个节点。 注意处理翻转的起始节点为head的情况。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *reverseBetween(ListNode *head, int m, int n) { 12 if (head == NULL) return NULL; 13 14 ListNode *pre_node = NULL, *mid_first = head; 15 for (int i = 1; i < m; i++) { 16 pre_node = mid_first; 17 mid_first = mid_first->next; 18 } 19 ListNode *last = mid_first->next, *pre = mid_first; 20 for (int i = m; i < n; i++) { 21 ListNode *tmp = last->next; 22 last->next = pre; 23 pre = last; 24 last = tmp; 25 } 26 if (pre_node == NULL) { 27 head = pre; 28 mid_first->next = last; 29 } else { 30 pre_node->next = pre; 31 mid_first->next = last; 32 } 33 return head; 34 } 35 };
LeetCode OJ - Reverse Linked List II,布布扣,bubuko.com
LeetCode OJ - Reverse Linked List II
标签:c style class blog code java
原文地址:http://www.cnblogs.com/dongguangqing/p/3775024.html