【LeetCode刷题记录】H-Index

题目

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.

解答

这题是求N篇论文的H指数。刚开始，没去看维基百科里的详细解释，所以没发现有已有的算法思路。折腾了快一个小时，才勉强AC过。不仅代码啰嗦，而且逻辑性不强。为了通过某些边界条件，修修补补，算是得到了教训。

我的思路：

用C++里algorithm的sort()对N篇论文的index升序排序（其实应该降序排序的）。

设最终的H-index为hIndex，hIndex的值范围在[0, N]之间。外层循环为hIndex的每一个可能值，从0到N。内层循环为升序排序后的每篇论文的index。使用smallCount表示引用次数小于hIndex的论文数，noMoreCount表示引用次数小于等于hIndex的论文数（后来一想这么设计完全是多此一举）。所以，内层循环终止的条件就是smallCount或者noMoreCount等于N减去当前可能的hIndex。这样，找到可能的每个hIndex，保存最大的。

AC代码：

class Solution {
public:
int hIndex(vector<int> &citations) {
int n = citations.size();
if (n == 0) {
return 0;
}

// sort in ascending order
sort(citations.begin(), citations.end());

// try each possible hindex from 0 to N
int hIndex = 0, hIndexMax = 0;

for (hIndex = 0; hIndex <= n; hIndex++) {
int smallNum = 0; // papers with citations less than h
int noMoreNum = 0; // papers with citations no more than h
vector<int>::iterator it;
//cout << "hindex : " << hIndex << endl;
for (it = citations.begin(); it != citations.end(); ++it) {
if (*it < hIndex) {
smallNum++;
noMoreNum++;
}
else if (*it == hIndex) {
noMoreNum++;
}
else {
//cout << "small num " << smallNum << " noMoreNum " << noMoreNum << endl;
if ((smallNum <= n - hIndex) || (noMoreNum <= n - hIndex)) {
hIndexMax = hIndexMax > hIndex ? hIndexMax : hIndex;
break;
}
}
}
if ((smallNum <= n - hIndex) || (noMoreNum <= n - hIndex)) {
hIndexMax = hIndexMax > hIndex ? hIndexMax : hIndex;
}
}

return hIndexMax;
}
};

标准解法：

wiki上直接给出了算法，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。我也就没多想，直接按照上面的方法写出了代码。

AC代码：

class Solution {
public:

static bool myfunction(int i, int j) {
return (i > j);
}

int hIndex(vector<int>& citations) {
sort(citations.begin(), citations.end(), myfunction);

for (int i = 0; i < citations.size(); i++)  {
if (i >= citations[i]) {
return i;
}
}
return citations.size();
}
};

感想

在得到一种基本的解决方法后，需要想想有没有简洁的方法。