两个较短阵列中的较长阵列 [英] Longer array out of two shorter arrays
问题描述
我的问题是：有两个单维数组，更长[M]和
更短[N]，每个数组都以升序排列订购。我们需要
从两个中构建一个新数组。这就是我写的：
#include< stdio.h>
#define M 8
#define N 5
main（）
{
int big [M] = {1,2,5,8,10,23,45,56};
int small [N] = {3,7,11,12,100};
int new [M + N];
int i = 0， k = 0，l = 0;
while（l<（N + M））
{
if（small [k]< big [i]）
{
new [l] = small [k];
k ++;
l ++;
}
else
{
new [l] = big [i ];
i ++;
l ++;
}
}
for（l = 0; l<（M + N）; l ++）
printf（"％d"，new [l]）;
printf （" \ n"）;
}
输出结果为：
[Mike @ localhost演习] $ ./153
1 2 3 5 7 8 10 11 12 23 45 56 8
为什么最后一个号码错了？
感谢您的关注！
./ 153
1 2 3 5 7 8 10 11 12 23 45 56 8
为什么最后一个号码错了？
感谢您的关注！
Mik0b0说：
大家好，
我的问题是：有两个单维数组，更长[M]和
更短[N]，每个数组按升序排列。我们需要
从两个中构建一个新数组。这就是我写的：
#include< stdio.h>
#define M 8
#define N 5
main（）
{
int big [M] = {1,2,5,8,10,23,45,56};
int small [N] = {3,7,11,12,100};
int new [M + N];
int i = 0， k = 0，l = 0;
while（l<（N + M））
{
if（small [k]< big [i]）
你不检查以确保k< N或者i< M.一旦k == N，你需要
停止加载''small''并复制剩下的''big''，除非i == M.一旦我
== M，你需要停止加载''big''，并复制剩下的''small''，
除非k == N.
解决这个问题，我认为你的问题会消失。
-
理查德希思菲尔德
" Usenet is一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上述域名中， - www。
Mik0b0写道：
更短[N]，每个数组按升序排列。我们需要
从两个中构建一个新数组。这就是我写的：
#include< stdio.h>
#define M 8
#define N 5
main（）
int main（无效）
< snip code>
>
输出结果为：
[Mike @ localhost演习]
Hallo everybody,
my problem is: there are two single-dimension arrays, longer[M] and
shorter[N], every array is organized in ascending order. We need to
build a new array out of two. This is what I wrote:
#include<stdio.h>
#define M 8
#define N 5
main()
{
int big[M]={1,2,5,8,10,23,45,56};
int small[N]={3,7,11,12,100};
int new[M+N];
int i=0,k=0,l=0;
while(l<(N+M))
{
if(small[k]<big[i])
{
new[l]=small[k];
k++;
l++;
}
else
{
new[l]=big[i];
i++;
l++;
}
}
for(l=0;l<(M+N);l++)
printf("%d ",new[l]);
printf("\n");
}
The output is:
[Mike@localhost drills]$ ./153
1 2 3 5 7 8 10 11 12 23 45 56 8
Why is the last number wrong?
Thanks for your attention!
./153
1 2 3 5 7 8 10 11 12 23 45 56 8
Why is the last number wrong?
Thanks for your attention!
Mik0b0 said:
Hallo everybody,
my problem is: there are two single-dimension arrays, longer[M] and
shorter[N], every array is organized in ascending order. We need to
build a new array out of two. This is what I wrote:
#include<stdio.h>
#define M 8
#define N 5
main()
{
int big[M]={1,2,5,8,10,23,45,56};
int small[N]={3,7,11,12,100};
int new[M+N];
int i=0,k=0,l=0;
while(l<(N+M))
{
if(small[k]<big[i])You don''t check to ensure that k < N or that i < M. Once k == N, you need to
stop loading from ''small'' and copy the rest of ''big'', unless i == M. Once i
== M, you need to stop loading from ''big'', and copy the rest of ''small'',
unless k == N.
Fix that, and I reckon your problem will vanish.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Mik0b0 wrote:Hallo everybody,
my problem is: there are two single-dimension arrays, longer[M] and
shorter[N], every array is organized in ascending order. We need to
build a new array out of two. This is what I wrote:
#include<stdio.h>
#define M 8
#define N 5
main()int main(void)
<snip code>>
The output is:
[Mike@localhost drills]
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