# H-Index

### 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.

### 分析​

H-Index 的含义是，如果一个人发表的所有论文中，有h篇论文分别被引用了至少h次，那么他的 H-Index 就是h

### 代码 1 全排序​

// H-Index// Time complexity: O(nlogn), Space complexity: O(1)public class Solution {    public int hIndex(int[] citations) {        Arrays.sort(citations);        reverse(citations);        for (int i = 0; i < citations.length; ++i) {            if (i + 1 == citations[i]) return i+1;            if (i + 1 > citations[i]) return i;        }        return citations.length;    }    private static void reverse(int[] nums) {        int left = 0;        int right = nums.length - 1;        while (left < right) {            final int tmp = nums[left];            nums[left] = nums[right];            nums[right] = tmp;            ++left;            --right;        }    }}

### 代码 2 计数排序​

// H-Index// Time complexity: O(n), Space complexity: O(n)public class Solution {    public int hIndex(int[] citations) {        final int n = citations.length + 1;        final int[] histogram = new int[n+1];        for (int x : citations) {            ++histogram[x > n ? n : x];        }        int sum = 0; // current number of papers        for (int i = n; i > 0; --i) {            sum += histogram[i];            if (sum >= i) {                return i;            }        }        return 0;    }}