problem:
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.
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题意:翻转单链表第m~n个范围的结点
thinking:
O(n)算法
code:
/** * 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) return NULL; ListNode *q = NULL; ListNode *p = head; for(int i = 0; i < m - 1; i++) { q = p; p = p->next; } ListNode *end = p; ListNode *pPre = p; p = p->next; for(int i = m + 1; i <= n; i++) { ListNode *pNext = p->next; p->next = pPre; pPre = p; p = pNext; } end->next = p; if (q) q->next = pPre; else head = pPre; return head; } };
时间: 2024-11-06 14:55:09