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."
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.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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，去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();
}
};
类似题目：
[LeetCode] 275. H-Index II H指数 II
