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:
Sort first then scan: O(NlogN) time, O(1)space
use another array: O(N) time, O(N) space
我们额外使用一个大小为N+1的数组stats。stats[i]表示有多少文章被引用了i次，这里如果一篇文章引用大于N次，我们就将其当为N次，因为H指数不会超过文章的总数。为了构建这个数组，我们需要先将整个文献引用数组遍历一遍，对相应的格子加一。统计完后，我们从N向1开始遍历这个统计数组。如果遍历到某一个引用次数时，大于或等于该引用次数的文章数量，大于引用次数本身时，我们可以认为这是H指数。之所以不用再向下找，因为我们要取最大的H指数。那如何求大于或等于某个引用次数的文章数量呢？我们可以用一个变量，从高引用次的文章数累加下来。因为我们知道，如果有x篇文章的引用大于等于3次，那引用大于等于2次的文章数量一定是x加上引用次数等于2次的文章数量。
1 public class Solution { 2 public int hIndex(int[] citations) { 3 int N = citations.length; 4 int[] summary = new int[N+1]; 5 for (int m : citations) { 6 if (m <= N) summary[m]++; 7 else summary[N]++; 8 } 9 int sum = 0; 10 for (int j=summary.length-1; j>=0; j--) { 11 sum += summary[j]; //sum means # of papers that has citation greater than or equal to j 12 if (sum >= j) return j; 13 } 14 return 0; 15 } 16 }
原文：http://www.cnblogs.com/EdwardLiu/p/5074786.html