问题：

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 : "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.

解决：

【题意】有h篇论文的引用数目大于等于h，其余引用小于h。

① 排序法。先将数组排序，我们就可以知道对于某个引用数，有多少文献的引用数大于这个数。对于引用数citations[i]，大于该引用数文献的数量是citations.length - i，而当前的H-Index则是Math.min(citations[i], citations.length - i)，我们将这个当前的H指数和全局最大的H指数来比较，得到最大H指数。

1)

class Solution { //4ms

    public int hIndex(int[] citations) {         Arrays.sort(citations);         int h = 0;//全局最大的H指数         for (int i = 0;i < citations.length;i ++){             int curH = Math.min(citations[i],citations.length - i);//得到当前的H指数             if (curH > h){                 h = curH;             }         }         return h;     } }

2)

class Solution{//2ms

    public int hIndex(int[] citations) {         Arrays.sort(citations);         for (int i = 0; i < citations.length; i ++) {             if (citations[i] >= citations.length - i) {                 return citations.length - i;             }         }         return 0;     } }

② 数组映射法。我们额外使用一个大小为N+1的数组dp。dp[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。

1)

class Solution { //1ms

    public int hIndex(int[] citations) {         int len = citations.length;         int[] dp = new int[len + 1];         for (int i = 0;i < len;i ++){//统计各个引用次数对应多少篇文章             int index = citations[i] <= len ? citations[i] : len;             dp[index] += 1;         }         int count = 0;         for (int i = len;i > 0;i --){             count+= dp[i];             if (count >= i){                 return i;             }         }         return 0;     } }

2)

class Solution { //1ms

    public int hIndex(int[] citations) {         int[] dp = new int[citations.length + 1];         for (int n : citations){             if (n >= citations.length){                 dp[citations.length] ++;             }else{                 dp[n] ++;             }         }         int count = 0;         for (int i = citations.length;i > 0;i --){             count += dp[i];             if (count >= i){                 return i;             }         }         return 0;     } }