
I spent lots of time on understanding what H is. In fact just image the vector like a bar and the value of each citations[i] is the height of each factor.

Definition: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."

Now you use the H from 0 to n (included) to test all of the factors in the vector. You should find one number of H which makes that that some certain citations[k], the n  k == citations[k], where the k is the value of our expected H.

So binary search is applied for Time O(lg n).
class Solution {
public:
int hIndex(vector<int>& citations) {
if(citations.empty()) return 0;
int n = citations.size();
int a = 0, b = n;// because the index range is [0,n].
while(a < b){
int i = a + (b  a)/2;
if(citations[i] > n  i){
b = i;
}
else if(citations[i] < n  i){
a = i + 1;
}
else if(citations[i] == n  i){
return n  i;
}
}
return n  a;
}
};