LeetCode 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 thanh 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.

题意：给出一个数组，求这样的一个数h,数组中的h个数的值大于等于h,如果有多个，取最大的。

思路：从大到小排序，判断当前元素组是否大于其下标，如果是，继续判断，直到当前元素值小于或等于下标

代码如下：

class Solution
{
    public int hIndex(int[] citations)
    {
        Comparator cmp = new Comparator() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o2 - o1;
            }
        };

        Integer[] tmp = new Integer[citations.length];
        for (int i = 0; i < tmp.length; i++)
        {
            tmp[i] = new Integer(citations[i]);
        }

        Arrays.sort(tmp, cmp);

        int i = 0;
        for (i = 0; i < citations.length; i++)
        {
            citations[i] = tmp[i].intValue();
            if (citations[i] <= i)
            {
                break;
            }
        }

        return i;
    }
}

方法二：用计数来达到排序，如果元素值大于数组长度，就将对应的count[len]加1，然后从尾到 头计算大于当前下标的数目，如果和大于当前下标，说明当前下标就是所求的值 。

代码如下：

class Solution
{
    public int hIndex(int[] citations)
    {
        int n = citations.length;
        int[] count = new int[n + 1];

        for (int i = 0; i < n; i++)
        {
            if (citations[i] >= n) count[n]++;
            else count[citations[i]]++;
        }

        int ans = 0;
        for (int i = n; i >= 0; i--)
        {
            ans += count[i];
            if (ans >= i) return i;
        }

        return ans;
    }
}