https://leetcode.com/problems/h-index/

Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.

According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the otherN − h papers have no more than h citations each."

For example, given citations = [3, 0, 6, 1, 5], which means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3citations each, his h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

Hint:

  1. An easy approach is to sort the array first.
  2. What are the possible values of h-index?
  3. A faster approach is to use extra space.
class Solution {
public:
int getIdx(vector<int>& vec, int val) {
if(vec.size() <= ) return -;
int low = , high = vec.size()-;
while(low < high) {
int mid = low + (high-low)/;
if(vec[mid] < val) low = mid+;
else high = mid;
}
return low;
}
int hIndex(vector<int>& citations) {
if(citations.size() == ) return ;
if(citations.size() == ) {
if(citations[] == ) return ;
return ;
}
sort(citations.begin(), citations.end());
if(citations[citations.size()-] == ) return ; int res = INT_MIN;
int low = , high = citations.size();
while(low < high) {
int mid = low + (high - low)/;
if(mid <= citations.size()-getIdx(citations, mid)) {
low = mid;
res = max(res, mid);
}
else high = mid;
if(high - low <= ) break;
} if(low <= citations.size()-getIdx(citations, low)) {
res = max(res, low);
}
if(high <= citations.size()-getIdx(citations, high)) {
res = max(res, high);
}
return res;
}
};

leetcode 263: H-Index

class Solution {
public:
int getIdx(vector<int>& vec, int val) {
if(vec.size() <= ) return -;
int low = , high = vec.size()-;
while(low < high) {
int mid = low + (high-low)/;
if(vec[mid] < val) low = mid+;
else high = mid;
}
return low;
}
int hIndex(vector<int>& citations) {
if(citations.size() == ) return ;
if(citations.size() == ) {
if(citations[] == ) return ;
return ;
}
//sort(citations.begin(), citations.end());
if(citations[citations.size()-] == ) return ; int res = INT_MIN;
int low = , high = citations.size();
while(low < high) {
int mid = low + (high - low)/;
if(mid <= citations.size()-getIdx(citations, mid)) {
low = mid;
res = max(res, mid);
}
else high = mid;
if(high - low <= ) break;
} if(low <= citations.size()-getIdx(citations, low)) {
res = max(res, low);
}
if(high <= citations.size()-getIdx(citations, high)) {
res = max(res, high);
}
return res;
}
};

leetcode 264: H-Index II

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