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- [网站分类]1.网站首页原创Java技术区(对首页文章的要求: 原创、高质量、经过认真思考并精心写作。BlogJava管理团队会对首页的文章进行管理。) (0)
最新评论
-
haohao-xuexi02:
很不错哦。
O'Reilly cos上传组件的使用(1/3) - 上传文件 -
yoin528:
useUnicode=true&charact ...
[原]向MySQL数据库插入Blob数据的问题 -
xiaoqing20:
下载来看看!呵呵
[原]Struts2类型转换 -
xiaoqing20:
[原]Struts2类型转换
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文章来源:http://wintys.blog.51cto.com/425414/100688
主元素.zip
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一、主元素问题
<!--[if !supportLists]--> 设T[0..n-1]是n个元素的数组。对任一元素x,设S(x)={i|T[i]=x}。当|S(x)|>n/2时,称x为T的主元素。
1) 如果T中元素存在序关系,按分治策略设计并实现一个线性时间算法,确定T[0..n-1]是否有一个主元素。
2) 若T中元素不存在序关系,只能测试任意两个元素是否相等,试设计并实现一个O(nlogn)有效算法,确定T是否有一个主元素。进一步,能找到一个线性时间算法吗?
注:实现的算法要求列出足够的实验结果。
1) 基于分治法的线性时间求主元素算法
<!--[if !supportLists]-->■
<!--[endif]-->算法思想
中位数:数列排序后位于最中间的那个数。如果一个数列有主元素,那么必然是其中位数。求一个数列有没有主元素,只要看中位数是不是主元素。
找中位数的方法:选择一个元素作为划分起点,然后用快速排序的方法将小于它的移动到左边,大于它的移动到右边。这样就将元素划分为两个部分。此时,划分元素所在位置为k。如果k>n/2,那么继续用同样的方法在左边部分找;如果k<n/2就在右边部分找;k=n/2就找到了中位元素。
根据快速排序的思想,可以在平均时间复杂度为O(n)的时间内找出一个数列的中位数。然后再用O(n)的时间检查它是否是主元素。
<!--[if !supportLists]-->■
<!--[endif]-->算法实现
对应的Java程序在MajorElement.java中
----------------------------------------------------------------------------------------
判断是否是主元素的伪代码:
master(A):
len
← length[A]
median
←
randomizedSelect(A , 0 , n - 1 , n/2); ▹求中位数
cnt
← 0
▹计算中位数出现次数
for
i ← 0 to len – 1
do if A[i] = median
then cnt ← cnt + 1
if
cnt > n/2
then print "主元素:" +median + "出现次数:"
+ cnt
else print "无主元素"
----------------------------------------------------------------------------------------
找一个序列中第k大的数伪代码
randomizedSelect(A , p , q , k):
r
←
randomizedPartition (p , q)
▹找出划分元素r
if
r = k
then return A[r]
else if r > k
then randomizedSelect(A , p , r – 1, k)
else randomizedSelect(A , r + 1 , q , k)
----------------------------------------------------------------------------------------
实现随机划分的伪代码:
randomizedPartition(A , p , q ):
rand ← random(p , q)
rand ← random(p , q)
exchange
A[rand] ↔A[q]
return
partition(p , q)
----------------------------------------------------------------------------------------
基于快速排序思想的划分伪代码:
partition(A , p , q ):
pivot
← A[q]
i
← p – 1
for
j ← p to q – 1
do
if A[j] <= pivot
then i ← i + 1
exchange
A[i] ↔ A[j]
exchange
A[i + 1] ↔ A[q]
return
i + 1
----------------------------------------------------------------------------------------
<!--[if !supportLists]-->■
<!--[endif]-->时间复杂度分析
master()中求中位数可以在平均时间复杂度为O(n)的时间内完成,检查中位数是否是主元素耗时O(n),所以时间复杂度为O(n)。
2) 无序关系时求主元素的O(nlgn)的算法
<!--[if !supportLists]-->■
<!--[endif]-->算法思想
若T 中存在主元素,则将T 分为两部分后,T 的主元素也必为两部分中至少一部分的主元素,因此可用分治法。
将元素划分为两部分,递归地检查两部分有无主元素。算法如下:
a. 若T 只含一个元素,则此元素就是主元素,返回此数。
b. 将T 分为两部分T1 和T2(二者元素个数相等或只差一个),分别递归调用此方法求其主元素m1 和m2。
c. 若m1 和m2 都存在且相等,则这个数就是T 的主元素,返回此数。
d. 若m1 和m2 都存在且不等,则分别检查这两个数是否为T 的主元素,若有则返回此数,若无则返回空值。
e. 若m1 和m2 只有一个存在,则检查这个数是否为T 的主元素,若是则返回此数,若否就返回空值。
f. 若m1 和m2 都不存在,则T 无主元素,返回空值。
<!--[if !supportLists]-->■
<!--[endif]-->算法实现
相应的Java程序在MasterElement.java中
-----------------------------------------------------------------------------------------
O(nlgn)的算法伪代码:
▹求T[p..q]中的主元素。返回主元素及其出现次数或空(表示无主元素)
CheckMaster(T , p , q):
if
p ← q
then
return T[p] and 1
len
← q – p + 1
r
← p + len / 2
a
and numa ← CheckMaster(T , p , r – 1)
b
and numb ← CheckMaster(T , r , q)
if
a = NIL and b = NIL
then
return NIL
if
a = NIL and b ≠ NIL
then
return CheckAnotherPart(T , len , p , r – 1 , b , numb)
if
a ≠ NIL and b = NIL
then
return CheckAnotherPart(T , len , r , q , a , numa)
if
a ≠ NIL and b ≠ NIL
then
if a = b
then numa ← numa + numb
return a and numa
else
re ← CheckAnotherPart(T , len , p , r – 1 , b ,numb)
if re ≠ NIL
then
return re
else
return CheckAnotherPart(T, len, r, q, a, numa)
-----------------------------------------------------------------------------------------
▹检查候选主元素是否是主元素
CheckAnotherPart(T , len , p , q , c ,
numc):
numc
← CheckNum(T , p , q , c) + numc
if
num > len/2
then
return c and numc
else
return NIL
-----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
▹计算T[p..q]中element出现的次数
CheckNum( T , p , q , element):
cnt
← 0
for
i ← p to q
do
if T[i] = element
then
cnt ← cnt + 1
return
cnt
----------------------------------------------------------------------------------------
<!--[if !supportLists]-->■
<!--[endif]-->时间复杂度分析
T(n)=2T(n/2)+n
所以时间复杂度为O(nlgn)
3)无序关系时求主元素的O(n)算法
<!--[if !supportLists]-->■
<!--[endif]-->算法思想
在一个集合中,删除两个不同的数,则集合的主元素保持不变。根据这个原理,可以很快求出主元素。
<!--[if !supportLists]-->■
<!--[endif]-->算法实现
-------------------------------------------------------------------------------------
相应的Java程序在MainElement.java中
master(A):
n
← length[A]
count
← 1
seed
← A[0]
▹找候选主元素
for
i ← 1 to n – 1
do
if A[i] = seed
then count ← count + 1
else if count > 0
then count ← count – 1
else seed ← A[i]
▹查找候选主元素是否是主元素
count
← 0
for
i ← 0 to n – 1
do if A[i] = seed
then count ← count + 1
if
count > n/2
then return seed and count
else return NIL
-------------------------------------------------------------------------------------
<!--[if !supportLists]-->■
<!--[endif]-->时间复杂度分析
时间复杂度为O(n)
4)算法测试
对前面三个求主元素算法,使用测试数据进行测试:
测试数组
|
结果
|
0,0,1,1,0,8,1,1,1
|
主元素:1出现次数:5
|
13,17,26,3,5,2,17,3
|
无主元素
|
1,2,3,4,5,6,7,8
|
无主元素
|
1,0,0,1,0,2,0
|
主元素:0 出现次数:4
|
1,3,4,1,2,1,1,4,0,1,1,1
|
主元素:1 出现次数:7
|
0,1,1
|
主元素:1 出现次数:2
|
13,17,26,3,5,2,17,3,3,3,3,3,3
|
主元素:3 出现次数:7
|
100,58
|
无主元素
|
597
|
主元素:597 出现次数:1
|
6,1,2,2,2,3,5
|
无主元素
|
7,7,7,7,7,7
|
主元素7 出现次数:6
|
5,9,45,96,77,28,13
|
无主元素
|
文章来源:http://wintys.blog.51cto.com/425414/100688
主元素.zip
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