0274. H-Index (M)

H-Index (M)

题目

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."

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, 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, her h-index is 3.

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

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题意

设一个研究者共有n篇论文，如果其中有h篇，这h篇中每一篇都被引用过至少h次，而剩余的n-h篇中每一篇被引用的次数都不超过h，则称h为这个研究者的h指数。求一个研究者h指数的最大值。

思路

h的范围很明显是 0-n，所以只要从n往回找到第一个满足的h，就是答案。因此可以这样处理：先将数组按升序排序，从头遍历数组，对于每一个i，length - i即为从i到最后一个元素的长度len，如果citations[i] >= len，说明后面len个论文的引用次数都大于等于len，且前面所有i个论文的引用次数必不大于len (证明如下)，因此len就是要找的h指数。

证明：假设在i处找到了满足的len，那么有 citations[i] > len (等于的情况无需证明)，如果 citations[i - 1] > len，一定有 citations[i - 1] >= len + 1 (因为是整数)，那么说明在i - 1处就应该已经找到了满足的len值，但实际情况并不是，所以 citation[i - 1] <= len，证明前i个论文的引用次数一定都不超过len，满足题目要求。

也可以不用排序：记citations.length为n，建一个长度为n + 1的辅助数组aux，aux[i]表示引用次数等于i的论文篇数，aux[n]表示引用次数大于等于n的论文篇数。遍历citations，如果citations[i] > n，则aux[n]++，否则aux[citations[i]]++。再从后向前遍历aux数组，统计当前和是否大于等于当前下标index，是的话说明index就是所要求的的h指数。

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代码实现

Java

排序

class Solution {
    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;
    }
}

不排序

class Solution {
    public int hIndex(int[] citations) {
        int len = citations.length;
        int[] aux = new int[len + 1];
        for (int i = 0; i < len; i++) {
            if (citations[i] > len) {
                aux[len]++;
            } else {
                aux[citations[i]]++;
            }
        }
        int sum = 0;
        for (int i = aux.length - 1; i >= 0; i--) {
            sum += aux[i];
            if (sum >= i) {
                return i;
            }
        }
        return 0;
    }
}