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 thanh 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 

3

 citations
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:
An easy approach is to sort the array first.
What are the possible values of h-index?
A faster approach is to use extra space.

 Solution 1 Sort
First, sort the array, and there are n - i elements(including citations[i]) on the right-hand side of citations[i], so there are
n - i elements >= citations[i], let h = n - i, if citations[i] >= n - 1, there are must be n - i elements that >= n - i, so n - i is what we need, because we want the maximum of h, so we start i from 0.

http://www.cnblogs.com/co0oder/p/5296417.html

public int hIndex(int[] citations) {
Arrays.sort(citations);
int n = citations.length;
for (int i = 0; i < n; i++)
if (citations[i] >= n - i)
return n - i;
return 0;
}

Solution2 O(N) without sort

//https://leetcode.com/discuss/55952/my-o-n-time-solution-use-java
public static int hIndex2(int[] citations) {
int length = citations.length;
if (length == 0) {
return 0;
}

int[] array2 = new int[length + 1];
for (int i = 0; i < length; i++) {
if (citations[i] > length) {
array2[length] += 1;
} else {
array2[citations[i]] += 1;
}
}
int t = 0;
for (int i = length; i >= 0; i--) {
t = t + array2[i];
if (t >= i) {
return i;
}
}
return 0;
}