C语言 c语言如何判断给定数组是否按降序排序
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How to find whether the given array is sorted in descending order in c programming
提问by Anish Kumar
I need to find out whether the given array is sorted in descending order or not...
I got the output but in portal it shows as Wrong Answer.
我需要找出给定的数组是否按降序排序...
我得到了输出，但在门户中它显示为错误答案。
This is my code.
这是我的代码。
#include<stdio.h>
int main()
{
int n,a[15],i,k=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{ scanf("%d",&a[i]);}
for(i=1;i<=n;i++)
{
if(a[i]<a[i+1])
k++;
}
if(k==0)
printf("yes");
else
printf("no");
return 0;
}
Help me to figure it out...
帮我想办法...
回答by Arun A S
Arrays are indexed from 0to size - 1. So if you have
数组的索引从0到size - 1。所以如果你有
int array[15];
the elements are a[0]to a[14]. But in your code, you are starting from a[1]to a[15]and might try to access a[15], which is unsafe memory and will cause problems.
的元素是a[0]对a[14]。但是在您的代码中，您从a[1]to开始a[15]并可能尝试访问a[15]，这是不安全的内存并会导致问题。
So you should first change your for loop.
所以你应该首先改变你的 for 循环。
You should change it to
你应该把它改成
for( i = 0 ; i < n ; i++ )
{
scanf("%d",&a[i]);
}
for( i = 0 ; i < n - 1 ; i++ )
{
if( a[i] < a[i+1] )
{
k++;
break;
}
}
In the second for loop, you should loop till i < n - 1because otherwise, in
在第二个 for 循环中，你应该循环直到i < n - 1因为否则，在
if(a[i]<a[i+1])
when i = n, you will try to access the n + 1th with element with a[i+1], which might be out of bounds.
当 时i = n，您将尝试访问n + 1带有元素的th a[i+1]，这可能会越界。
You can also just exit from the loop once you find out that the array is not in descending order to save time.
您也可以在发现数组不是按降序排列时退出循环以节省时间。
You must also be certain that a[15]is enough to store all the values ( that is, the number of values given as input should not exceed 15, check the problem statement to ensure this )
您还必须确定a[15]足以存储所有值（即作为输入给出的值数量不应超过 15，请检查问题陈述以确保这一点）
回答by Himanshu
you should start array index from 0in both for loop.
您应该从0两个 for 循环中开始数组索引。
Use this
用这个
for(i=0;i<n;i++)
in second for loop you should use
在第二个 for 循环中你应该使用
for(i=0;i<n-1;i++) // you need to compare up to second last element with last, so run loop upto `i<n-1`
for example if you enter n = 5loop will work for 0to 4
例如，如果您输入n = 5循环将适用0于4
It is good to use break;if value of kincreases
break;如果值k增加，最好使用
for(i=1;i<=n;i++)
{
if(a[i]<a[i+1]){
k++;
break; // if k increments break the loop.
}
}
回答by Vlad from Moscow
Here you are.:)
这个给你。：）
#include <stdio.h>
#define N 15
int main(void)
{
int a[N];
int i, n;
printf( "Enter number of elements in the array (not greater than %d: ", N );
scanf( "%d", &n );
if ( N < n ) n = N;
printf( "Enter %d elements of the array: ", n );
for ( i = 0; i < n; i++ ) scanf( "%d", &a[i] );
i = 0;
while ( i++ < n && !( a[i-1] < a[i] ) );
if ( i == n ) puts( "The array is sorted in the descending order" );
else puts( "The array is not sorted in the descending order" );
return 0;
}
The program output might look like
程序输出可能看起来像
Enter number of elements in the array (not greater than 15: 15
Enter 15 elements of the array: 10 10 9 9 9 8 7 6 5 5 4 3 2 1 0
The array is sorted in the descending order
As for you code then this loop
至于你的代码然后这个循环
for(i=1;i<=n;i++)
{
if(a[i]<a[i+1])
k++;
}
is unsafe because using index i+1in expression a[i+1]you can access memory beyond the array.
Also you have to check whether the entered value of n is less than 15 that is the size of the array because you use the range of indices [0, n].
And in the first loop the index has to start from 0. Otherwise the first element of the array will not be initialized.
是不安全的，因为i+1在表达式中使用索引a[i+1]可以访问数组之外的内存。此外，您必须检查输入的 n 值是否小于数组大小的 15，因为您使用了索引范围[0, n]。在第一个循环中，索引必须从 0 开始。否则数组的第一个元素将不会被初始化。
回答by leicar
First, you define array a[15]:
首先，定义数组 a[15]：
int n,a[15],i,k=0;
and you use variable n to control its size
并且您使用变量 n 来控制其大小
scanf("%d",&n);
for(i=1;i<=n;i++)
{ scanf("%d",&a[i]);}
then you need to keep n <= 15.
那么你需要保持 n <= 15。
p.s., array starts with '1' in Matlab but '0' in C.
ps，数组在Matlab中以'1'开头，在C中以'0'开头。
回答by NUMBART
I can't agree less about what others have said regarding array indexing, the problem in not knowing whether a[i+1] exists or not, and using of break. I would add though when all the numbers in this input array are equal it will still pass as decreasing in your code. If that is a problem you can add another if statement checking whether at all it decreases anywhere and use another variable say change for checking whether it gets satisfied.
对于其他人对数组索引、不知道 a[i+1] 是否存在以及使用 break 的问题，我完全同意。我想补充一点，当这个输入数组中的所有数字都相等时，它仍然会在你的代码中作为递减传递。如果这是一个问题，您可以添加另一个 if 语句，检查它是否在任何地方减少，并使用另一个变量 say change 来检查它是否得到满足。
int change = 0;
for(i=0;i<n;i++)
{
if(a[i]<a[i+1])
k++;
if(a[i]>a[i+1])
change++;
}
if(k==0 && change > 1)
printf("yes");
else
printf("no");
Check with the problem statement whether it allows such cases as descending. If it doesn't go with this otherwise its fine.
检查问题陈述是否允许降序等情况。如果它不符合这个，否则它很好。
