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一个软件公司面试题

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一个软件公司面试题

发表于：2007-07-12 17:40:26 楼主

两个text（text1,text2),和一个command1
我text1输入一些字符包括数字==
然后单击command1
要在text2显示
（text1取得每个字符的ascii，从小到大依次排序）
比如在text1输入aa1，单击command1
text2显示出 1aa

发表于：2007-07-12 19:52:301楼得分:0

private sub command1_click（)
if len（text1.text) <=0 then
exit sub
end if

dim nsize as integer,i as integer,j as integer,bchanged as boolean
nsize = len（text1.text)
dim atmp（nsize-1) as integer

for i=0 to ubound（atmp)
' 赋值
atmp（i) = asc（mid（text1.text,i,1))

next i

'冒泡排序
for i = 0 to ubound（atmp) -1
bchanged = false
for j=0 to ubound（atmp)-i
if（atmp（j) > atmp（j+1)) then
dim ntmp as integer
ntmp=atmp（j+1)
atmp（j+1) = atmp（j)
atmp（j) = ntmp
bchanged = true
end if
next j
if bchanged = false then
exit for
end if
next i

' 在text2中显示处理好以后的数据，按asc从小到大输出;
for i = 0 to ubound（atmp)
text2.text = text2.text & chr（atmp（i))
next i
end sub

没vb6在手上，没测试

发表于：2007-07-12 21:00:082楼得分:0

private sub command1_click（)
dim t1%
t1 = len（text1.text)
for i = 1 to t1
text2.text = mid$（text1.text, i, 1) + text2.text
next
end sub

发表于：2007-07-13 10:25:023楼得分:0

to: zhzboy（) （ ) 信誉：100 blog 加为好友 2007-7-12 19:52:30 得分: 0

你的代码好象不对哦
测试不通过....不过也谢谢你
--------------------------
to: bfblang（) （ ) 信誉：100 blog 加为好友 2007-7-12 21:00:08 得分: 0
你的代码．．．．．．．．．．．

发表于：2007-07-13 10:32:314楼得分:0

private sub command1_click（)
dim szstr as string
dim inti as integer
dim icnt as integer
dim arry（) as string
szstr = text2.text
inti = len（szstr)

for icnt = 0 to len（szstr) - 1
redim preserve arry（icnt)
arry（icnt) = asc（mid（szstr, icnt + 1, 1))
next icnt

call quicksort（arry（), 0, ubound（arry))
szstr = " "
for icnt = 0 to ubound（arry)
szstr = szstr & chr（arry（icnt))

next icnt

text2.text = szstr

end sub

sub quicksort（marry（) as string, l as integer, r as integer)
dim i as integer, j as integer, x as integer, y as integer
i = l
j = r
x = marry（（l + r) / 2)
while （i <= j)
while （marry（i) < x and i < r)
i = i + 1
wend

while （x < marry（j) and j > l)
j = j - 1
wend

if （i <= j) then
y = marry（i)
marry（i) = marry（j)
marry（j) = y
i = i + 1
j = j - 1
end if
wend

if （l < j) then call quicksort（marry（), l, j)
if （i < r) then call quicksort（marry（), i, r)

end sub

发表于：2007-07-13 10:59:005楼得分:0

我的咋了？简单实用，用最少的代码完成任务不久行了嘛？

发表于：2007-07-13 11:04:116楼得分:0

private sub command1_click（)
dim i%
for i = 1 to len（text1.text)
text2.text = mid$（text1.text, i, 1) + text2.text
next
end sub
太复杂了,我一直想学

发表于：2007-07-13 13:27:167楼得分:0

在form中添加listbox,listbox的sorted属性设为true
private sub command1_click（)
text2.text = " "
list1.clear
dim i as integer
for i = 1 to len（text1.text)
list1.additem mid（text1.text, i, 1)
next i
for i = 0 to list1.listcount - 1
text2.text = text2.text & list1.list（i)
next i

end sub不知是这意思吗？

发表于：2007-07-13 13:51:138楼得分:0

private sub command1_click（)
dim my_n（), my_w（), my_flg（)
dim i as long, j as long, cc as long
redim my_n（1 to len（text1.text))
redim my_w（1 to len（text1.text))
redim my_flg（1 to len（text1.text))
text2.text = " "
for i = 1 to len（text1.text)
my_n（i) = asc（mid（text1.text, i, 1))
my_flg（i) = true
next
cc = 1
i = 1
g_next: do while i <= len（text1.text)
if my_flg（i) = false then i = i + 1: goto g_next
min_n = my_n（i)
rr = i
for j = 1 to len（text1.text)
if min_n > my_n（j) and my_flg（j) = true then rr = j: min_n = my_n（j)
next
my_w（cc) = min_n
my_flg（rr) = false
cc = cc + 1
loop
for i = 1 to len（text1.text)
text2.text = text2.text & chr（my_w（i))
next
end sub

发表于：2007-07-13 15:19:249楼得分:0

private sub command1_click（)
dim s
text2.text = " "
dim t1
t1 = text1.text
dim i
for i = 1 to len（t1)
s = mid（t1, i, 1)
dim ipos
ipos = i
dim j
for j = i + 1 to len（t1)
if mid（t1, j, 1) < s then
s = mid（t1, j, 1)
ipos = j
end if
next
dim tt1
tt1 = left（t1, i - 1) & s
if ipos > i then
tt1 = tt1 & mid（t1, i + 1, ipos - i - 1) & mid（t1, i, 1)
end if
tt1 = tt1 & right（t1, len（t1) - ipos)
t1 = tt1
text2.text = text2.text & s
next

end sub

发表于：2007-07-13 15:37:5510楼得分:0

'这个方法也可以

private sub command1_click（)

text2.text = resetarr（text1.text)

end sub

function resetarr（byval s as string) as string
dim arr（)
redim preserve arr（len（s) - 1)
dim i
for i = 1 to len（s)
arr（i - 1) = mid（s, i, 1)
next

for i = 0 to ubound（arr)
s = arr（i)
dim ipos
ipos = i
dim j
for j = i + 1 to ubound（arr)
if arr（j) < s then
s = arr（j)
ipos = j
end if
next
arr（ipos) = arr（i)
arr（i) = s
next
s = " "
for i = 0 to ubound（arr)
s = s & arr（i)
next
resetarr = s
end function

发表于：2007-07-13 16:26:1411楼得分:0

如果只包含英文字符和数字，还可以这样
不需要排序，但是效率不高
private sub command1_click（)
dim arr（1 to 128) as long
dim i as long, nasc as long
dim s as string
s = text1.text
for i = 1 to len（s)
nasc = asc（mid（s, i, 1))
arr（nasc) = arr（nasc) + 1
next
s = " "
for i = 1 to 128
s = s & string（arr（i), chr（i))
next
text2.text = s
end sub

发表于：2007-07-13 19:49:3712楼得分:0

private sub command1_click（)
text2.text = " "
if len（text1.text) <= 0 then
exit sub
end if

dim nsize as integer, i as integer, j as integer, bchanged as boolean, atmp（) as integer
nsize = len（text1.text)
redim atmp（nsize - 1) as integer

for i = 0 to ubound（atmp)
'赋值
atmp（i) = asc（mid（text1.text, i + 1, 1))

next i

'冒泡排序
for i = 1 to ubound（atmp)
bchanged = false
for j = 0 to ubound（atmp) - i
if （atmp（j) > atmp（j + 1)) then
dim ntmp as integer
ntmp = atmp（j + 1)
atmp（j + 1) = atmp（j)
atmp（j) = ntmp
bchanged = true
end if
next j
if bchanged = false then
exit for
end if
next i

'在text2中显示处理好以后的数据，按asc从小到大输出;
for i = 0 to ubound（atmp)
text2.text = text2.text & chr（atmp（i))
next i
end sub

测试了,改了,这个可行

发表于：2007-07-14 21:04:2613楼得分:0

这个软件公司很强。要求算法层面的技术人员。

哈哈哈，一定是开发后台程序的。

虽然冒泡最基础，但是可以发明他的人智商也是万里挑一的。

发表于：2007-07-15 02:07:4914楼得分:0

好象有个内部函数可以直接输入相反的东西哦.

发表于：2007-07-15 10:32:2015楼得分:0

学习了

发表于：2007-08-04 16:03:0816楼得分:0

谢谢各位了...
zhzboy（) （ ) 信誉：100

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