[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:
Input:citations = [3,0,6,1,5]Output: 3 Explanation:[3,0,6,1,5]means the researcher has5papers in total and each of them had received3, 0, 6, 1, 5citations respectively. Since the researcher has3papers with at least3citations each and the remaining two with no more than3citations each, her h-index is3.
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();
}
};
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[LeetCode] 275. H-Index II H指数 II
