Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
Summary: Just a practice of binary search.
vector<int> searchRange(int A[], int n, int target) {
vector<int> range(,-);
int start = ;
int end = n - ;
while(start <= end) {
int median = start + (end - start + ) / ;
if(A[median] > target) {
end = median - ;
}else if (A[median] < target) {
start = median + ;
}else { //equals
//go right
int i = median + ;
for(; i <= end; i ++) {
if(A[i] != target){
range[] = i - ;
break;
}
}
if(range[] == -)
range[] = i - ;
//go left
i = median - ;
for(; i >= start; i --) {
if(A[i] != target){
range[] = i + ;
break;
}
}
if(range[] == -)
range[] = i + ;
break;
}
}
return range;
}