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

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

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

H指數（H index）是一個混合量化指標，可用於評估研究人員的學術產出數量與學術產出水平

可以按照如下方法確定某人的H指數：

將其發表的所有SCI論文按被引次數從高到低排序；

解法1: 先將數組排序，T：O(nlogn)， S：O(1)。然後對於每個引用次數，比較大於該引用次數的文章，去引用次數和文章數的最小值，即 Math.min(citations.length-i, citations[i])，並更新 level，取最大值。排好序之後可以用二分查找進行遍歷，這樣速度會更快，可見：275. H-Index II H指數 II

解法2: Counting sort，T：O(n)， S：O(n)。使用一個大小為 n+1 的數組count統計引用數，對於count[i]表示的是引用數為 i 的文章數量。從後往前遍歷數組，當滿足 count[i] >= i 時，i 就是 h 因子，返回即可，否則返回0。

為什麽要從後面開始遍歷？ 為什麽 count[i] >= i 時就返回？

一方面引用數引用數大於 i-1 的數量是i-1及之後的累加，必須從後往前遍歷。另一方面，h 因子要求盡可能取最大值，而 h 因子最可能出現最大值的地方在後面，往前值只會越來越小，能盡快返回就盡快返回，所以一遇到 count[i] >= i 就返回。參考：Code_Granker

Java:

public class Solution {  
    public int hIndex(int[] citations) {  
        Arrays.sort(citations);  
        int level = 0;  
        for(int i = 0; i < citations.length; i++)  
            level = Math.max(level,Math.min(citations.length - i,citations[i]));  
        return level;  
    }  
} 　　

Java:

public class Solution {  
    public int hIndex(int[] citations) {  
        int n = citations.length;  
        int[] count = new int[n + 1];  
        for(int c : citations)  
            if(c >= n) count[n]++;  //當引用數大於等於 n 時，都計入 count[n]中  
            else count[c]++;  
        for(int i = n; i > 0; i--) {  //從後面開始遍歷  
            if(count[i] >= i) return i;  
            count[i-1] += count[i];  //引用數大於 i-1 的數量是i-1及之後的累加  
        }  
        return 0;  
    }  
}　

Python: Counting sort.

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        n = len(citations);
        count = [0] * (n + 1)
        for x in citations:
            # Put all x >= n in the same bucket.
            if x >= n:
                count[n] += 1
            else:
                count[x] += 1

        h = 0
        for i in reversed(xrange(0, n + 1)):
            h += count[i]
            if h >= i:
                return i
        return h　

Python: T: O(nlogn) O: O(1)

class Solution2(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        citations.sort(reverse=True)
        h = 0
        for x in citations:
            if x >= h + 1:
                h += 1
            else:
                break
        return h

Python:　T: O(nlogn) O: O(n)　

class Solution3(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        return sum(x >= i + 1 for i, x in enumerate(sorted(citations, reverse=True))) 

Python:

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        if not citations: return 0
        return max([min(i + 1, c) for i, c in enumerate(sorted(citations, reverse=True))])　　

C++:

class Solution {
public:
    int hIndex(vector<int>& citations) {
        sort(citations.begin(), citations.end(), greater<int>());
        for (int i = 0; i < citations.size(); ++i) {
            if (i >= citations[i]) return i;
        }
        return citations.size();
    }
};　

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