LeetCode 25 - Reverse Nodes in k-Group

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

public ListNode reverseKGroup(ListNode head, int k) {
    if(head == null || head.next == null || k<=1) return head;
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode node = dummy;
    int i = 0;
    while(head != null) {
        i++;
        if(i%k == 0) {
            node = reverse(node, head.next);
            head = node.next;
        } else {
            head = head.next;
        }
    }

    return dummy.next;
}

private ListNode reverse(ListNode preHead, ListNode end) {
    ListNode tail = preHead.next;
    ListNode walker = tail.next;
    while(walker != end) {
        tail.next = walker.next;
        walker.next = preHead.next;
        preHead.next = walker;
        walker = tail.next;
    }
    return tail;
}

猜你喜欢

转载自yuanhsh.iteye.com/blog/2196878