# Question

Formatted question description: https://leetcode.ca/all/274.html

 274	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 other N − h papers have no more than h citations each."

Example:

Input: citations = [3,0,6,1,5]
Output: 3

Explanation: [3,0,6,1,5] 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 3 citations each, her h-index is 3.

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



# Algorithm

1. Sort all SCI papers published by him in descending order of the number of citations;
2. Search the sorted list from front to back until the serial number of a paper is greater than the number of times the paper has been cited.

The obtained serial number minus one is the H index

# Code

Java

• 
public class H_Index {
class Solution {
public int hIndex(int[] citations) {

if (citations == null || citations.length == 0) {
return 0;
}

// @note
Arrays.sort(citations); // 6,5,3,1,0
for (int i = 0; i < citations.length / 2; ++i) {
int temp = citations[i];
citations[i] = citations[citations.length - i - 1];
citations[citations.length - i - 1] = temp;
}

for (int i = 0; i < citations.length; ++i) {
if (citations[i] <= i) { // @note: <i then return i+1 will not work
return i;
}
}

return citations.length;

}
}
}


• // OJ: https://leetcode.com/problems/h-index/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int hIndex(vector<int>& A) {
sort(begin(A), end(A));
int L = 0, R = A.size() - 1, N = A.size();
while (L <= R) {
int M = (L + R) / 2;
if (A[M] < N - M) L = M + 1;
else R = M - 1;
}
return N - L;
}
};

• class Solution(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
n = len(citations)
dp = [0] * (n + 1)
for c in citations:
if c > n:
dp[n] += 1
else:
dp[c] += 1

total = 0
for i in reversed(range(1, len(dp))):
total += dp[i]
if total >= i:
return i
return 0