leetcode274~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.

下面的两个方法是基于计数排序的，第二个方法是对第一个方法的改进。
都是O(n)的时间复杂度。只不过第二个方法的空间复杂度更小。

public class H_Index {    //基于计数排序的思想，但是这种比较浪费空间    //使用计数排序的前提：存在k，使得数组里所有元素的值都不大于N，且元素的值都是非负。     public int hIndex2(int[] citations) {         if(citations.length==0 || citations==null) return 0;        Arrays.sort(citations);        int[] arr = new int[citations[citations.length-1]+1];        for(int i=0;i<citations.length;i++) {            arr[citations[i]] ++;        }        int sum=0;        for(int i=arr.length-1;i>=0;i--) {            sum = sum+arr[i];            if(sum>=i) return i;        }        return 0;    }    //基于计数排序，对其进行变形     public int hIndex(int[] citations) {         int[] arr = new int[citations.length+1]; //这里直接按照长度进行分配空间 因为返回的index的最大值肯定小于数组的长度         for(int i=0;i<citations.length;i++) {             if(citations[i]>=citations.length) {                 arr[citations.length] ++;             } else {                 arr[citations[i]] ++;             }         }        int sum=0;        for(int i=citations.length;i>=0;i--) {            sum += arr[i];            if(sum>=i) return i;         }        return 0;     }}