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.
/** * 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(m==n) return head; int num = 1; ListNode *h = head; ListNode *Pm = head,*Pm_pre=NULL; while(num != m){ Pm_pre = Pm; Pm = Pm->next; num++; }//end while ListNode *p1 = Pm,*p2 = p1->next,*p3 = NULL; while(num!=n){ p3 = p2->next; p2->next = p1; p1 = p2; p2 = p3; num++; }//end while if(Pm_pre!=NULL){ Pm_pre->next = p1; Pm->next = p2; }else{ h = p1; Pm->next = p2; } return h; } };