[LeetCode]59. 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."

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.

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

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解法1：对输入数组按引用次数从高到低排序，然后扫描一遍数组，找到第一篇“被引用次数小于它在数组中出现的序号”的论文，其序号即是H-Index。

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

解法2：设计Map<int,int>，保存下某个引用次数和论文篇数的映射。扫描一遍数组，若当前论文引用次数为x，则将[0,x]间所有的引用次数的论文篇数加1。然后再扫描一遍Map，找出论文篇数超过这个引用次数的引用次数，即是H-Index。

class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
map<int, int> index;
for (int i = 0; i < n; ++i) {
int idx = citations[i];
for (int j = 0; j <= idx; ++j)
++index[j];
}
int hIndex = 0;
for (map<int, int>::iterator iter = index.begin(); iter != index.end(); ++iter) {
if (iter->second >= iter->first)
hIndex = max(hIndex, iter->first);
}
return hIndex;
}
};