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。

 

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