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, 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, his h-index is 3.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
分析
需要看懂 H-index 的定义
如果作者有 X 篇文章的引用，每个都不少于 X，那么其 H-index 是 X
可以看出，最大 H-index 是文章总数
所以建立一个长度为 citations.size() + 1 的数组A，A[i] 代表引用次数大于 i 的文章数。
遍历一遍引用数据，然后从数组 A 的尾部向前查找，直到发现 总数 大于 i，即为所求
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | class Solution {public: int hIndex(vector<int>& citations) { sort(citations.begin(), citations.end()); int len = citations.size(); vector<int> A(len + 1, 0); int h_count = 0, h_index = 0; for(int i = 0; i < len; ++i){ citations[i] >= len ? ++A[len]: ++A[citations[i]]; } for(int i = len; i >=0; --i){ h_count += A[i]; if(h_count >= i){ h_index = i; break; } } return h_index; }}; |