H指数的计算,这道题是典型的排序思路,博主第一次使用一般的排序方法写,后来发现整数这种用基数排序更合适,同时需要考虑很多特例,博主贴了这两份代码。由于测试案例的原因,基数排序速度上的优势没有体现出来。
原题
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.
一般的排序解法
这里主要是对题意的理解,排序之后一个循环解决问题。
class Solution { public: int hIndex_old(vector<int>& citations) { sort(citations.rbegin(), citations.rend()); const int len = citations.size(); if (len == 0){ return 0; } for (int i = 0; i < len; ++i){ if (citations[i] < i+1){ return i; } } return len; } };
基数排序解法
因为都是整数,所以比较适合基数排序。可能是由于数据长度过小的原因,基数排序的优势没有体现出来。在使用基数排序的时候,因为涉及到数组下标,所以要特别注意边界条件。比如等号的选取,大于长度时单独存放(否则[0, 0, 0]这种就会出错等等),要考虑很多特例,没有上面解法直白。
class Solution { public: int hIndex(vector<int>& citations){ const int len = citations.size(); if (len == 0){ return 0; } vector<int> count(len+1); for (int i = 0; i < len; ++i){ if (citations[i] >= len){ count[len]++; } else if (citations[i] >= 0){ count[citations[i]]++; } else{ //ciatations[i] < 0 count[0]++; // report error; } } int total_cite = 0; for (int i = len; i >= 0; --i){ total_cite += count[i]; if (total_cite >= i){ return i; } } return 0; } };
掌握好基数排序的话,还是推荐基数排序方法。