英文:
Fixed Point (Value equal to index) in a given array using Linear Search - Logic question
问题
以下是您要翻译的部分：
"Can someone provide an explanation why using this method below returns only the first matching value within the array? For example if My array element is 0, 1, 2, 3 and 4. It will return 0, instead of 4. Every value within the array matches to its index 4, should using the for loop return 4 after the last iteration is completed?
static int linearSearch(int arr[], int n)
{
int i;
for(i = 0; i < n; i++)
{
if(arr[i] == i)
return i;
}
/* If no fixed point present
then return -1 */
return -1;
}
//main function"
英文:
Can someone provide an explanation why using this method below returns only the first matching value within the array? For example if My array element is 0, 1, 2, 3 and 4. It will return 0, instead of 4. Every value within the array matches to its index 4, should using the for loop return 4 after the last iteration is completed?
static int linearSearch(int arr[], int n)
{
int i;
for(i = 0; i < n; i++)
{
if(arr[i] == i)
return i;
}
/* If no fixed point present
then return -1 */
return -1;
}
//main function
答案1
得分: 3
这是因为在你的 if 语句中，你立即调用了 return，而你的方法应该返回一个单独的 int。如果你想收集所有的固定点，你可以修改你的方法如下：
static List<Integer> linearSearch(int arr[], int n)
{
List<Integer> fixedPoints = new ArrayList<>();
for(int i = 0; i < n; i++)
{
if(arr[i] == i)
fixedPoints.add(i);
}
return fixedPoints;
}
英文:
It's because in your if you immediately call return, and your method is supposed to return a single int. If you want to collect all fixed points, you can modify your method like this:
static List<Integer> linearSearch(int arr[], int n)
{
List<Integer> fixedPoints = new ArrayList<>();
for(int i = 0; i < n; i++)
{
if(arr[i] == i)
fixedPoints.add(i);
}
return fixedPoints;
}
答案2
得分: 2
这将返回4，因为它具有每个i匹配到数组中的值的更新索引。
static int linearSearch(int arr[], int n)
{
int i, index = -1;
for (i = 0; i < n; i++)
{
if (arr[i] == i)
index = i;
}
/* 如果没有固定点存在
则返回-1 */
return index;
}
// 主函数
英文:
This will return 4 as it has an updated index for each i matching to its value in an array.
static int linearSearch(int arr[], int n)
{
int i,index = -1;
for(i = 0; i < n; i++)
{
if(arr[i] == i)
index = i;
}
/* If no fixed point present
then return -1 */
return index;
}
//main function
答案3
得分: 1
方法在达到return语句时返回给调用它的代码。根据文档：
从方法返回一个值
当方法
- 完成了方法中的所有语句，
- 达到了return语句，或者
- 抛出了异常，
会返回给调用它的代码，以先发生的情况为准。
语句 if(arr[i] == i) 在for循环的第一步返回true：
i=0
arr[0]=0
然后执行了 return i;。
在这种情况下，数组的每个值都与它的索引匹配，但在最后的迭代完成后，使用for循环不会返回4，因为最后的迭代是for循环的第一次迭代，只针对i=0。i 的值永远不会达到4。
英文:
> Q: Can someone provide an explanation why using this method below
> returns only the first matching value within the array?
A method returns to the code that invoked it when reaches a return statement.
From doc:
> Returning a Value from a Method
>
> A method returns to the code that invoked it when it
>
> - completes all the statements in the method,
> - reaches a return statement, or
> - throws an exception,
>
> whichever occurs first.
The statement if(arr[i] == i) returns true at the first step of the for loop:
i=0
arr[0]=0
Then the line return i; executed.
> Q: Every value within the array matches to its index 4, should using
> the for loop return 4 after the last iteration is completed?
In this condition the last iteration is the first iteration of the for loop. It runs only for i=0. The value of i never reaches to 4.
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