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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 {//边遍历边旋转,目标数目节点旋转完成后,与前后未变部分进行对接 10 public: 11 ListNode* reverseBetween(ListNode* head, int m, int n) { 12 ListNode *ans=head; 13 ListNode *pri=head; 14 int count=n-m; 15 if(m==1) 16 ans=NULL; 17 else 18 ans=head; 19 if(m==1) 20 { 21 ListNode *rev=pri; 22 ListNode *acc=rev->next; 23 ListNode *revlast=rev; 24 while(count--) 25 { 26 ListNode *temp=acc; 27 acc=acc->next; 28 temp->next=rev; 29 rev=temp; 30 } 31 revlast->next=acc; 32 ans=rev; 33 } 34 else 35 { 36 m--; 37 while(--m) 38 pri=pri->next; 39 40 ListNode *rev=pri->next; 41 ListNode *acc=rev->next; 42 ListNode *revlast=rev; 43 while(count--) 44 { 45 ListNode *temp=acc; 46 acc=acc->next; 47 temp->next=rev; 48 rev=temp; 49 } 50 revlast->next=acc; 51 pri->next=rev; 52 } 53 return ans; 54 } 55 };
leetcode92 Reverse Linked List II
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原文地址:http://www.cnblogs.com/jsir2016bky/p/5106008.html