Leetcodes Solutions 25 Reverse Nodes in k-Group

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Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked 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

思路1

常規做法，關鍵在於對for循環中交換k個節點的理解

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode *reverseKGroup(ListNode *head, int k){ if(head == NULL || k == 1) return head; int num = 0; ListNode *preheader = new ListNode(-1); preheader->next = head; ListNode *cur = preheader, *nex, *pre = preheader; while(cur = cur->next) num++; while(num>=k){ cur = pre->next; nex = cur->next; for(int i = 1; i < k; ++i){ cur->next = nex->next; nex->next = pre->next; pre->next = nex; nex = cur->next; } pre = cur; num -= k; } return preheader->next; }};

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