[LeetCode] 025. Reverse Nodes in k-Group (Hard) (C++/Java)
来源：程序员人生 发布时间：2015-03-14 10:01:02 阅读次数：3073次
索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github:
https://github.com/illuz/leetcode
025. Reverse Nodes in k-Group (Hard)
链接：
题目：https://oj.leetcode.com/problems/reverse-nodes-in-k-group/
代码(github)：https://github.com/illuz/leetcode
题意：
把1个链表每 k 个分为1组，每组内进行翻转。
只能用常数级的空间。
分析：
这题比较考验链表的操作，用递归做会比较方便，先找到下1组的节点，把本组反转后再递归处理后面的节点。
代码：
C++:
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (!head || !(head->next) || k < 2)
return head;
// count k nodes
ListNode *nextgp = head;
for (int i = 0; i < k; i++)
if (nextgp)
nextgp = nextgp->next;
else
return head;
// reverse
ListNode *prev = NULL, *cur = head, *next = NULL;
while (cur != nextgp) {
next = cur->next;
if (prev)
cur->next = prev;
else
cur->next = reverseKGroup(nextgp, k);
prev = cur;
cur = next;
}
return prev;
}
};
Java:
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode cur = head;
int cnt = 0;
// get next group
while (cur != null && cnt != k) {
cur = cur.next;
cnt++;
}
if (cnt == k) {
cur = reverseKGroup(cur, k);
// reverse
while (0 <= --cnt) {
ListNode tmp = head.next;
head.next = cur;
cur = head;
head = tmp;
}
head = cur;
}
return head;
}
}
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