[LeetCode] Reverse Nodes in k-Group

来源：互联网发布：工地扬尘标准数据编辑：程序博客网时间：2024/05/17 04:32

在上一个题的基础上改动了一下，因为内存受限，所以每次都从头寻找指定的节点，并单独实现了一个函数。

可读性比较差，估计过几天就看不懂自己写的了。不知道别人有什么更好的办法。等有空的时候好好研究一下链表的操作。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode reverseKGroup(ListNode head, int k) {        ListNode resultHead = new ListNode(0);        resultHead.next = head;        ListNode tempHead = new ListNode(-1);        tempHead.next = resultHead;        ListNode tempEnd = new ListNode(-1);        ListNode lastNode = getNextNode(tempHead, k + 1);        if (lastNode == null) {            return resultHead.next;        }        while (lastNode != null) {            ListNode theNextStart = lastNode.next;            tempEnd.next = lastNode;            ListNode firstNode = tempHead.next.next;            if (firstNode.next != null) {                tempHead.next.next = lastNode;                for (int j = k - 1; j > 0; --j) {                    ListNode n = getNextNode(firstNode, j - 1);                    tempEnd.next.next = n;                    tempEnd.next = tempEnd.next.next;                }                tempHead.next = firstNode;                firstNode.next = theNextStart;            } else {                break;            }            lastNode = getNextNode(tempHead, k + 1);        }        return resultHead.next;            }            public ListNode getNextNode(ListNode head, int n) {        ListNode result = new ListNode(0);        result.next = head;        for (int i = 0; i < n; ++i) {            result.next = result.next.next;            if (result.next == null) {                return null;            }        }        return result.next;    }}

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