[LeetCode] 274. H-Index H指数

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:

citations = [3,0,6,1,5]
[3,0,6,1,5] 
5
3, 0, 6, 1, 5
3
3
3
3

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

H指数（H index）是一个混合量化指标，可用于评估研究人员的学术产出数量与学术产出水平

可以按照如下方法确定某人的H指数：

将其发表的所有SCI论文按被引次数从高到低排序；

从前往后查找排序后的列表，直到某篇论文的序号大于该论文被引次数。所得序号减一即为H指数。

解法1: 先将数组排序，T：O(nlogn)， S：O(1)。然后对于每个引用次数，比较大于该引用次数的文章，取引用次数和文章数的最小值，即 Math.min(citations.length-i, citations[i])，并更新 level，取最大值。排好序之后可以用二分查找进行遍历，这样速度会更快，可见：275. H-Index II H指数 II

解法2: Counting sort，T：O(n)， S：O(n)。使用一个大小为 n+1 的数组count统计引用数，对于count[i]表示的是引用数为 i 的文章数量。从后往前遍历数组，当满足 count[i] >= i 时，i 就是 h 因子，返回即可，否则返回0。

为什么要从后面开始遍历？为什么 count[i] >= i 时就返回？

一方面引用数引用数大于 i-1 的数量是i-1及之后的累加，必须从后往前遍历。另一方面，h 因子要求尽可能取最大值，而 h 因子最可能出现最大值的地方在后面，往前值只会越来越小，能尽快返回就尽快返回，所以一遇到 count[i] >= i 就返回。参考：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]++;  //当引用数大于等于 n 时，都计入 count[n]中  
            else count[c]++;  
        for(int i = n; i > 0; i--) {  //从后面开始遍历  
            if(count[i] >= i) return i;  
            count[i-1] += count[i];  //引用数大于 i-1 的数量是i-1及之后的累加  
        }  
        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();
    }
};　