[LeetCode] H-Index

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.

解题思路

类似于计数排序，略。

实现代码

C++：

// Runtime: 4 ms
class Solution {
public:
    int hIndex(vector<int>& citations) {
        int len = citations.size();
        vector<int> cnt(len + 1, 0);
        for_each(citations.begin(), citations.end(),
                 [len, &cnt](int n){ n >= len ? cnt[len]++ : cnt[n]++; });
        for (int i = len; i >= 0; i--)
        {
            cnt[i] = i == len ? cnt[i] : cnt[i] + cnt[i + 1];
            if (cnt[i] >= i)
            {
                return i;
            }
        }

        return 0;
    }
};

Java:

// Runtime: 1 ms
public class Solution {
    public int hIndex(int[] citations) {
        int len = citations.length;
        int cnt[] = new int[len + 1];
        for (int c : citations) {
            if (c >= len) {
                cnt[len]++;
            }
            else {
                cnt[c]++;
            }
        }

        for (int i = len; i >= 0; i--) {
            cnt[i] += i == len ? 0 : cnt[i + 1];
            if (cnt[i] >= i) {
                return i;
            }
        }

        return 0;
    }
}