Sero现在还能转么？转了还能保留号码到别的carrier么？# PDA - 掌中宝
b*n
1 楼
"Programming Pearls" has a binary search program that searches the first
occurrence of a certain value. I think it is a little confusing so I wrote
it in another way:
/* Function that returns the first occurrence of x.
* arr is input sorted array, arr_size is the number of elements in arr,
* x is the value that we are looking for.
* Returns index of the element, or -1 if element is not found. */
int first_occurrence(int* arr, int arr_size, int x)
{
int l = 0;
int u = arr_size - 1;
while (l + 1 < u) {
int m = l + (u - l) / 2;
if (arr[m] < x)
l = m + 1; // line 8: both "l = m" and "l = m + 1" will work here?
else
u = m;
}
if (arr[l] == x)
return l;
if (arr[u] == x)
return u;
return -1;
}
My question is on line 8 (see the comments on that line). I tested a few
cases and it seems that both "l = m" or "l = m + 1" will work. Can anybody
confirm that? Thanks.
occurrence of a certain value. I think it is a little confusing so I wrote
it in another way:
/* Function that returns the first occurrence of x.
* arr is input sorted array, arr_size is the number of elements in arr,
* x is the value that we are looking for.
* Returns index of the element, or -1 if element is not found. */
int first_occurrence(int* arr, int arr_size, int x)
{
int l = 0;
int u = arr_size - 1;
while (l + 1 < u) {
int m = l + (u - l) / 2;
if (arr[m] < x)
l = m + 1; // line 8: both "l = m" and "l = m + 1" will work here?
else
u = m;
}
if (arr[l] == x)
return l;
if (arr[u] == x)
return u;
return -1;
}
My question is on line 8 (see the comments on that line). I tested a few
cases and it seems that both "l = m" or "l = m + 1" will work. Can anybody
confirm that? Thanks.
