bsearch method
Assuming the list is sorted, locate the leftmost element exactly equal to x
.
If the element is found, the index of element is returned. If not found, -1 is returned.
Implementation
int bsearch(E x, {Comparator<E>? compare, int low = 0, int? high}) {
compare = argToComparator<E>(compare, null);
final i = this.bisectLeft(x, compare: compare, low: low, high: high);
if (i != this.length && compare(this[i], x) == 0) {
return i;
}
return -1;
}