14. Reverse Linked List II

Reverse Linked List II

Reverse a linked list from positionMToN. Do it in-place and in one-pass.

For example: Given1->2->3->4->5->NULL,M= 2 andN= 4,

Return1->4->3->2->5->NULL.

Note:GivenM,NSatisfy the following condition: 1 ≤M≤N≤Length of list.

 

Conclusion: It is actually a reverse linked list. It is only the middle part of the reversal. Note that the cursor of the first node is saved. If the first node is the header node, note that the tail node of the anti-Rotor String is changed to the header node.

/** * 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) { ListNode *preNode1, *node1, *node2; preNode1 = node1 = node2 = NULL; int cnt = 0; for(ListNode *p = head; p != NULL; p = p->next) { cnt++; if(cnt == m-1) preNode1 = p; // hidden: m > 1 if(cnt == m) node1 = p; if(cnt == n) { node2 = p; break; } } ListNode *tail = node2->next; // must take out as a tag. ListNode *pre = tail, *post; while(node1 != tail) { post = node1->next; node1->next = pre; pre = node1; node1 = post; } if(m > 1) { preNode1->next = node2; return head;} return node2; // node1 is the 1st node. } };

 

14. Reverse Linked List II