reverse-nodes-in-k-group

https://leetcode.com/problems/reverse-nodes-in-k-group/

https://leetcode.com/mockinterview/session/result/xjlzcwc/

链表分批倒置，还是有点绕的。用了三个临时变量来处理。需要细心，也需要掌握类似经验。

package com.company;import java.util.*;class ListNode {int val;ListNode next;ListNode(int x) { val = x; }
}class Solution {public ListNode reverseKGroup(ListNode head, int k) {if (k <= 1) {return head;}ListNode ret = new ListNode(0);ret.next = head;ListNode cur = ret;ListNode next = head;ListNode first = null;ListNode second = null;ListNode third = null;while (true) {int i=0;for (; i<k; i++) {if (next == null) {break;}next = next.next;}if (i < k) {break;}first = cur.next;second = first.next;for (i=0; i<k-1; i++) {third = second.next;second.next = first;first = second;second = third;}cur.next.next = next;third = cur.next;cur.next = first;cur = third;}return ret.next;}
}public class Main {public static void main(String[] args) throws InterruptedException {System.out.println("Hello!");Solution solution = new Solution();// Your Codec object will be instantiated and called as such:ListNode ln1 = new ListNode(1);ListNode ln2 = new ListNode(2);ListNode ln3 = new ListNode(3);ListNode ln4 = new ListNode(4);ListNode ln5 = new ListNode(5);ln1.next = ln2;ln2.next = ln3;ln3.next = ln4;ln4.next = ln5;ListNode ret = solution.reverseKGroup(ln1, 3);for (int i=0; i<5; i++) {System.out.printf("ret:%d\n", ret.val);ret = ret.next;}System.out.println();}}

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