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 leasth citations each, and the other N − h papers have no more than

 h citations each."

Example:

Input: citations = [3,0,6,1,5]Output: 3 
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than
 
 3 citations each, her h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

題意：給出一個n個元素的陣列，找出一個h，這個陣列有h個數大於等於h，剩下h個數小於等於h，輸出最大的h。

思路：

解法一：再定義一個數組，記錄每個元素出現了多少次，再對這個陣列求字尾和，然後從後向前遍歷，如果字尾和大於等於i，停止遍歷。時間複雜度O（N）。

C程式碼：

int hIndex(int* citations, int citationsSize) {
    int ans = 0,i;
    int pre[1005];
    memset(pre,0,sizeof(pre));
    for(i = 0; i < citationsSize; i++) {
        pre[citations[i]]++;
    }
    for(i = 1000; i >= 0; i--) {
        pre[i] = pre[i + 1] + pre[i];
        if(pre[i] >= i) {
            ans = i;
            break;
        }
    }
    return ans;
}
 

解法二：先對陣列排序，從後向前遍歷，如果此時列舉到的數大於等於已經列舉的數的個數，停止遍歷。時間複雜度O（NlogN）。

Java程式碼：

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