leetcode:Reverse Nodes in k-Group(以k为循环节反转链表)【面试算法题】
2013-12-02 23:08
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题目:
Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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:
For k = 2, you should return:
For k = 3, you should return:
题意:给定一个链表和一个数字k,以k为循环节,反转链表中的元素,如果剩下的长度不足k,则最后一部分不进行反转。
先计算链表的长度,长度除以循环节长度,统计要进行链表反转的次数。
每个循环节中的链表反转用三个指针去维护。
当操作是第一个循环节时,需要改变head指针的值。
之后每个循环节反转结束后用ph和pe两个指针去维护循环节之间的指针连接。
题解目录
Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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
题意:给定一个链表和一个数字k,以k为循环节,反转链表中的元素,如果剩下的长度不足k,则最后一部分不进行反转。
先计算链表的长度,长度除以循环节长度,统计要进行链表反转的次数。
每个循环节中的链表反转用三个指针去维护。
当操作是第一个循环节时,需要改变head指针的值。
之后每个循环节反转结束后用ph和pe两个指针去维护循环节之间的指针连接。
/** * 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) { ListNode *p1,*p2,*p3,*ph,*pe; if(head==NULL)return NULL; int n=0,i,j; p1=head; if(k==1)return head; while(p1) { n++; p1=p1->next; } p1=head; p2=p1->next; if(p2)p3=p2->next; for(i=0;i<n/k;++i) { for(j=1;j<k;++j) { p2->next=p1; p1=p2; p2=p3; if(p3)p3=p3->next; } if(i==0) { ph=head; ph->next=p2; pe=p2; head=p1; } else { ph->next=p1; ph=pe; ph->next=p2; pe=p2; } p1=p2; if(p2)p2=p2->next; if(p3)p3=p3->next; } return head; } }; // blog.csdn.net/havenoidea
题解目录
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