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76 changes: 76 additions & 0 deletions
solution/025.Reverse Nodes in k-Group/README.md
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| ## k个一组翻转链表 | ||
| ### 题目描述 | ||
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| 给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 | ||
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| k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。 | ||
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| 示例 : | ||
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| 给定这个链表:1->2->3->4->5 | ||
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| 当 k = 2 时,应当返回: 2->1->4->3->5 | ||
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| 当 k = 3 时,应当返回: 3->2->1->4->5 | ||
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| 说明 : | ||
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| 你的算法只能使用常数的额外空间。 | ||
| 你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换 | ||
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| ### 解法 | ||
| 1. 在 head 节点前增加一个头节点 reNode 使所有的翻转操作情况一致。 | ||
| 2. 维护一个 num 计数,指针 pNode 从 head 节点开始,每经过 k 个节点,进行一次 k 个节点的翻转 | ||
| 3. 将翻转后的 k 个节点与前后组的节点相连 | ||
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| ```java | ||
| /** | ||
| * Definition for singly-linked list. | ||
| * public class ListNode { | ||
| * int val; | ||
| * ListNode next; | ||
| * ListNode(int x) { val = x; } | ||
| * } | ||
| */ | ||
| class Solution { | ||
| public ListNode reverseKGroup(ListNode head, int k) { | ||
| if(head == null || k < 2) { | ||
| return head; | ||
| } | ||
| int num = 0; | ||
| ListNode pNode = head; | ||
| ListNode lastNode = new ListNode(0); | ||
| ListNode reNode = lastNode; | ||
| lastNode.next = head; | ||
| while (pNode != null) { | ||
| num++; | ||
| if(num >= k) { | ||
| num = 0; | ||
| ListNode tempNode = pNode.next; | ||
| reserver(lastNode.next, k); | ||
| // k 个节点的尾节点指向下一组的头节点 | ||
| lastNode.next.next = tempNode; | ||
| // 上一组的尾节点指向当前 k 个节点的头节点 | ||
| tempNode = lastNode.next; | ||
| lastNode.next = pNode; | ||
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| lastNode = tempNode; | ||
| pNode = lastNode.next; | ||
| } | ||
| else { | ||
| pNode = pNode.next; | ||
| } | ||
| } | ||
| return reNode.next; | ||
| } | ||
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| private ListNode reserver(ListNode node, int i) { | ||
| if(i <= 1 || node.next == null) { | ||
| return node; | ||
| } | ||
| ListNode lastNode = reserver(node.next, i - 1); | ||
| lastNode.next = node; | ||
| return node; | ||
| } | ||
| } | ||
| ``` |
49 changes: 49 additions & 0 deletions
solution/025.Reverse Nodes in k-Group/Solution.java
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,49 @@ | ||
| /** | ||
| * Definition for singly-linked list. | ||
| * public class ListNode { | ||
| * int val; | ||
| * ListNode next; | ||
| * ListNode(int x) { val = x; } | ||
| * } | ||
| */ | ||
| class Solution { | ||
| public ListNode reverseKGroup(ListNode head, int k) { | ||
| if(head == null || k < 2) { | ||
| return head; | ||
| } | ||
| int num = 0; | ||
| ListNode pNode = head; | ||
| ListNode lastNode = new ListNode(0); | ||
| ListNode reNode = lastNode; | ||
| lastNode.next = head; | ||
| while (pNode != null) { | ||
| num++; | ||
| if(num >= k) { | ||
| num = 0; | ||
| ListNode tempNode = pNode.next; | ||
| reserver(lastNode.next, k); | ||
| // k 个节点的尾节点指向下一组的头节点 | ||
| lastNode.next.next = tempNode; | ||
| // 上一组的尾节点指向当前 k 个节点的头节点 | ||
| tempNode = lastNode.next; | ||
| lastNode.next = pNode; | ||
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| lastNode = tempNode; | ||
| pNode = lastNode.next; | ||
| } | ||
| else { | ||
| pNode = pNode.next; | ||
| } | ||
| } | ||
| return reNode.next; | ||
| } | ||
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| private ListNode reserver(ListNode node, int i) { | ||
| if(i <= 1 || node.next == null) { | ||
| return node; | ||
| } | ||
| ListNode lastNode = reserver(node.next, i - 1); | ||
| lastNode.next = node; | ||
| return node; | ||
| } | ||
| } |
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