<LeetCode OJ> 61 / 86 Rotate List / Partition List

来源：互联网发布：为什么阿里云要备案编辑：程序博客网时间：2024/05/22 21:18

61 Rotate List

Total Accepted: 69038 Total Submissions: 300765 Difficulty: Medium

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

分析：

具体分析见代码，比较简单！

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *///1,首先获取链表的长度，再根据k计算正序的新链表头//2，建立连接class Solution {public:    ListNode* rotateRight(ListNode* head, int k) {        if(head==NULL || k==0)            return head;        int len=0;        ListNode* pmove=head;        ListNode* pend=NULL;        ListNode* pnewhead=NULL;        //1，获取链表长度        while(pmove)        {            len++;            pend=pmove;//记录尾节点            pmove=pmove->next;        }                //2，找到新头的位置        k=k%len;//取余        if(k==0)            return head;        int nlen=len-k;        pmove=head;        while(--nlen)            pmove=pmove->next;       //3，建立连接        pnewhead=pmove->next;        pmove->next=NULL;        pend->next=head;        return pnewhead;    }};

86 Partition List

Total Accepted: 67540 Total Submissions: 226814 Difficulty: Medium

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

分析：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* partition(ListNode* head, int x) {        if(head==NULL || head->next==NULL)            return head;        ListNode node1(0),node2(0);//看到这里就猛然醒悟！！！草...        ListNode* pleft=&node1;        ListNode* pright=&node2;        while(head)        {            if(head->val < x)            {                pleft->next=head;                pleft=head;            }            else            {                pright->next=head;                pright=head;            }            head=head->next;            }        pleft->next=node2.next;        pright->next=NULL;        return node1.next;    }};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/51422358

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895

2 0