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搜索关键字:partition by    ( 2821个结果
ROW_NUMBER()OVER(partition)的使用
select cardno,planname,ROW_NUMBER() over(partition by cardno,planname order by createtime asc) as rowfrom CardAddrBlack即:将cardno,planname看作一个分组,以他们内部的...
分类:其他好文   时间:2014-08-14 19:58:59    阅读次数:140
Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s....
分类:其他好文   时间:2014-08-14 16:47:18    阅读次数:218
Oracle 同名字段的该行数据按照创建时间最新的隐藏其他
1、需求,表 SYS_INFO 的 NAME 字段会重复,按照 创建时间CREATE_AT 字段,取最新一条,其他隐藏SELECT * FROM (SELECT T.*,ROW_NUMBER() OVER(PARTITION BY NAME ORDER BY CREATE_AT DESC) AS ....
分类:数据库   时间:2014-08-14 13:38:18    阅读次数:203
[LeetCode] Partition List
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 o...
分类:其他好文   时间:2014-08-14 00:56:47    阅读次数:199
leetcode 刷题之路 82 Partition List
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...
分类:其他好文   时间:2014-08-13 15:02:26    阅读次数:208
Palindrome Partitioning II
Given a string s, partition s such that every substring of the partition is a palindrome.Return the minimum cuts needed for a palindrome partitioning ...
分类:其他好文   时间:2014-08-13 12:45:46    阅读次数:166
Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome.Return all possible palindrome partitioning of s.For example,...
分类:其他好文   时间:2014-08-13 12:23:16    阅读次数:237
算法之快速排序
这里简单的介绍下快速排序的伪代码实现和示意图:QuickSort(A,p,r)if p<r q=Partition(A,p,r) QuickSort(A,p,q-1) QucikSort(A,q+1,r)算法的关键部位为Partition函数的实现,它实现了对数组A(p,r)的原址...
分类:其他好文   时间:2014-08-13 00:42:04    阅读次数:217
SqlServerDBA 常用代码
--查看partition的四个视图select * from sys.partition_functions--查看分区函数select * from sys.partition_parametersselect * from sys.partition_range_values--查看分区函数对...
分类:数据库   时间:2014-08-12 21:53:04    阅读次数:181
n张牌判断是否是顺子
将大小王看成0 排序 然后求出0的个数 如果0的个数大于差值且不为对子就返回true #include //n张牌判断是否是顺子 using namespace std; int partition(int *number,int start,int end){ int temp = number[start]; while(start < end){ while(start ...
分类:其他好文   时间:2014-08-12 13:38:24    阅读次数:142
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