2014年4月10日 星期四

[LeetCode] Reverse Nodes in k-Group

Problem:
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
Solution: O(n)
public class Solution {
   public ListNode reverseKGroup(ListNode head, int k) {
        int count = countNode(head);
        return reverseKGroupwithCount(head,k,count);
    }
     
    public ListNode reverseKGroupwithCount(ListNode head, int k,int count) {
        if(head == null || k <2 )
            return head;
            
        if(count < k)
            return head;
            
        ListNode p = head;
        ListNode c = head.next;
        for(int i=0; i<k-1;i++){
            ListNode tmp = c.next;
            c.next = p;
            p = c;
            c = tmp;
        }
        head.next = reverseKGroupwithCount(c,k,count-k);
        return p;
    }
    
    public int countNode(ListNode head){
        int ret = 0;
        while(head!=null){
            ret++;
            head= head.next;
        }
        return ret;
    }
}

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