H-Index

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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.

解法1：

要注意h的范围是在[0, N]之间。
只需要用一个array记录每一个数字出现的次数，对于数字大于等于N的记录在N
最后从后往前，累加所有的次数，当count >= i的时候说明就是我们要求的h-index
https://discuss.leetcode.com/topic/40765/java-bucket-sort-o-n-solution-with-detail-explanation/2
C++

1

Java

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public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null || citations.length == 0) {
            return 0;
        }
        Arrays.sort(citations);
        int max = 0;
        for (int i = 0; i < citations.length; i++) {
            int curr = Math.min(citations[i], citations.length - i);    // [1,4,5]
            max = Math.max(max, curr);
        }
        return max;
    }
}

解法2： Bucket Sort的思想

目标是找寻一个数i，使得比他引用次数多的paper的数量（包括和他一样的）要比i大就可以。统计每一个citation次数出现的次数，然后从大到小找到第一个满足条件的数字就可以了。

lang: java
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class Solution {
    public int hIndex(int[] citations) {
        if (citations == null || citations.length == 0) {
            return 0;
        }
        int n = citations.length;
        int[] buckets = new int[n + 1];
        for (int citation : citations) {
            if (citation >= n) {
                buckets[n]++;
            } else {
                buckets[citation]++;
            }
        }
        // move from back to front
        int count = 0;
        for (int i = n; i >= 0; i--) {
            count += buckets[i];
            if (count >= i) {
                return i;
            }
        }
        return 0;
    }
}