题目描述：
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.
本题是说，对于一个数组arr中的每一个数arr[i]，其中i∈[0,n-1] ，找到元素个数k，其中0<k<=n 。arr中有k个元素大于等于k，而arr中的其余n-k个元素均小于等于k。
思路：
对于数组arr，如果进行倒序排序再遍历的话，对于i和arr[i]关系，由于i是递增的，而arr[i]递减，只要找到第一个i>=arr[i]，那么后面第n-i个元素一定满足i>=arr[i]。因此，在遍历过程中，只要统计所有满足i<arr[i]的元素个数即可。
因此本题可以转换为：对数组arr进行倒序排序，统计index < arr[index]的个数。例如对于
3,5,7,1,2,3来说，排序后为: 7,5,3,3,2,1
0 < 7 => count ++
1 < 5 => count ++
2 < 3 => count ++
3 = 3
4 > 2
5 > 1
因此H-index 为count = 3
实现代码：
public class Solution {
public int HIndex(int[] citations) {
if(citations.Length == 0){
return 0;
}
var sorted = citations.OrderByDescending(x=>x).ToList();
var sum = 0;
for(var i = 0;i < sorted.Count; i++){
if(i < sorted[i]){
sum ++;
}
}
return sum;
}
}
LeetCode--H-Index
原文：http://blog.csdn.net/lan_liang/article/details/50145121
