leetcode題解-274. H-Index

題目：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.

本題是要求一個科學家的H_Index。也就是論文索引的一個名額。有h_index的定義可知，其值一定在0~citations範圍内。是以我們可以使用一個數組來儲存論文的索引次數。代碼入下，可以擊敗58%的使用者。

public static int hIndex(int[] citations) {
        int n = citations.length, tot=;
        //arr用來儲存每個索引次數論文的數量。arr[0]就是索引次數為0的論文數
        //其長度比citations大一，是因為arr[n]用來儲存索引次數大于n的論文數，因為索引次數大于n的文章一定是滿足h_index的文章。
        int[] arr = new int[n+];
        for (int i=; i<n; i++) {
        //周遊citations數組，并将論文的索引資訊儲存到arr數組中
            if (citations[i]>=n) arr[n]++;
            else arr[citations[i]]++;
        }
        //為了求最大的h_index，是以倒叙周遊arr，直到找到滿足條件的索引i并傳回。
        for (int i=n; i>=; i--) {
            tot += arr[i];
            if (tot>=i) return i;
        }
        return ;
    }
           

當然除此之外也可以先把數組進行排序，但相比上面方法的兩次循環而言效率反倒有所下降，主要是因為排序算法的效率是o(nlogn)，而上面代碼的效率是o(n)。代碼入下：

public int hIndex(int[] citations) {
        if (citations == null || citations.length ==) return;
        Arrays.sort(citations);
        int len = citations.length;
        for (int i =; i < citations.length; i++) {
            if (len <= citations[i])
                return len;
            else
                len--;
        }
        return len;
    }
           

此外，還可以采用排序之後在進行二分搜尋的方式，代碼入下：

public int hIndex(int[] citations) {
    Arrays.sort(citations);

    int n = citations.length;
    int i = , j = n - ;

    while (i <= j) {
        int k = (i + j) / ;
        int v = citations[k];
        int h = n - k;
        if (v >= h) {
            j = k - ;
        } else {
            i = k + ;
        }
    }

    return n - j - ;
}