[Leetcode 73] 92 Reverse Linked List II

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.

 

Analysis:

This is an extension problem of reverse a linked list. The first thing to do is to find the starting node of the sub-list to be reversed. Then call the reverse linked-list procedure and adjust the order of nodes. At last, link the new head with the m-1 node together to form the result linked-list. Note in the reverse procedure we connect the sub-list's last node with the n+1 node.

It's a one-pass and in place algorithm.

 

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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL || head->next == NULL)
            return head;
            
        ListNode dmy(-1), *p, *pm;
        dmy.next = head;
        p = &dmy;
        
        int cnt = 0;
        for (; cnt < m-1; cnt++) {
            p = p->next;
        }

        // now p points to m-1's node
        p->next = reverse(p->next, cnt+1, n);
        
        return dmy.next;
    }
    
private:
    ListNode *reverse(ListNode *head, int c, int n) {
        if (head == NULL || head->next == NULL)
            return head;
            
        ListNode *pre, *cur, *nxt, dmy(-1);
        dmy.next = head;
        
        pre = 0;
        cur = dmy.next;
        nxt = cur->next;
        
        while (c <= n) {
            nxt = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nxt;
            
            c++;
        }
        
        head->next = cur;
        return pre;
    }
};