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.

Hint:

An easy approach is to sort the array first. What are the possible values of h-index? A faster approach is to use extra space.

Tips

可以按照如下方法确定某人的H指数：

1、将其发表的所有SCI论文按被引次数从高到低排序；

O(nlogn)时间，O(1)空间。现将数组排序。然后从大到小遍历（traverse），一边计数一边比较计数与当前的引用数，直到计数大于引用数。

2、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。

只是用多余的一个空间记了个数，也是相当于排序了，其实和1是一样的。

O(n)时间，O(n)空间。

使用一个额外的数组，其下标为引用数，置为为具有该引用数的文章数量。注意，根据定义，H-index的上限不可能超过文章总数n。因此我们只需要额外开一个长度为n的数组即可。然后对新数组按引用数从大到小遍历，一边计数一边比较计数与当前的引用数，直到计数大于引用数。

Code:

方法一：

public class Solution {
    public int hIndex(int[] citations) {
        int len = citations.length;
        Arrays.sort(citations);
        for (int i = citations.length - 1; i >= 0; i--) {
            int j = citations.length - i - 1;
            if (j >= citations[i]) {
                return j;
            }
        }
        return citations.length;
    }
}

方法二；

public class Solution {
    public int hIndex(int[] citations) {
        int len = citations.length;
        if(citations == null || len == 0) return 0;
        int[] counts = new int[len + 1];
        for(int c : citations){
            if(c > len) counts[len]++;
            else counts[c]++;
        }
        if(counts[len] >= len) return len;
        for(int i = len - 1; i >= 0; i--){
            counts[i] += counts[i + 1];
            if(counts[i] >= i) return i;
        }
        return 0;
    }
}