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. 一开始想的是排序，之后从后向前遍历，这个位置的数大于长度与这个位置的差，说明符合条件，h++，最后返回h。代码如下： public class Solution { public int hIndex(int[] citations) { Arrays.sort(citations); int h = 0; for (int i = citations.length - 1; i >= 0; i --) { if (citations[i] >= (citations.length - i)) { h ++; } else { break; } } return h; } }Top Solution采用bucket sort，新建一个数组array[len + 1]，遍历citations[len]数组，大于len的话array[len] ++，否则array[citations[i]]++，之后从后向前遍历累加array[len + 1]，超过当前的位置值就说明有h篇论文的索引大于或者等于当前的索引数i。代码如下： public class Solution { public int hIndex(int[] citations) { int len = citations.length; if (len == 0) { return 0; } int[] array = new int[len + 1]; for (int i = 0; i < len; i ++) { if (citations[i] > len) { array[len] ++; } else { array[citations[i]] ++; } } int h = 0; for (int i = len; i >=0; i --) { h += array[i]; if (h >= i) { return i; } } return 0; } }