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.
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 (m == n) return head; 13 ListNode dummy(-1); 14 dummy.next = head; 15 ListNode *p = &dummy; 16 for (int i = 0; i < m - 1; ++i) { 17 p = p->next; 18 } 19 //p指向第m个结点的前一个结点 20 ListNode *t = p->next, *q = t->next; 21 //t指向已经reverse的部分的最后一个结点,q指向即将reverse的下一个结点 22 for (int i = m; i < n; ++i) { 23 t->next = q->next; 24 q->next = p->next; 25 p->next = q; 26 q = t->next; 27 } 28 return dummy.next; 29 } 30 };
一定要画图分析,小心指针操作。
【Leetcode】Reverse Linked List II,布布扣,bubuko.com
【Leetcode】Reverse Linked List II
原文:http://www.cnblogs.com/dengeven/p/3740211.html