LeetCode274. H-Index

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本文介绍了一种计算科研人员h指数的方法，h指数反映科研人员影响力，若科研人员有h篇论文至少被引用h次，其余论文不超过h次，则其h指数为h。文章提供了四种算法实现：暴力求解、二分查找、排序和桶排序。

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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.

思路：

这道题的解法有很多，以下提供四种作为参考：

解法1 暴力求解

我们从小到大对可能的答案进行判断：

class Solution {
public:
	int hIndex(vector<int>& citations) {
		int H = 0;
		for (H = citations.size(); H >= 0; H--) {
			if (validH(citations, H)) break;
		}
		return H;
	}
private:
	bool validH(const vector<int>& citations, int H) {
		int count = 0;
		for (int citation : citations) if (citation >= H) count++;
		return count >= H;
	}
};

解法2 二分查找

我们根据找到的符合条件的数的个数作为阈值进行二分查找：

class Solution {
public:
	int hIndex(vector<int>& citations) {
		int left = 0, right = citations.size();
		int mid = (left+right)/2;
		while (left < right) {
			if (validH(citations, mid) >= mid) left = mid + 1;
			else right = mid - 1;
			mid = (left + right) / 2;
		}
		int count = validH(citations, mid);
		if (count >= mid) return mid;
		else return mid - 1;
	}
private:
	int validH(const vector<int>& citations, int H) {
		int count = 0;
		for (int citation : citations) if (citation >= H) count++;
		return count;
	}
};

解法3 排序

我们将原序列排序后，找到第一个满足条件的值：

class Solution {
public:
	int hIndex(vector<int>& citations) {
		sort(citations.begin(), citations.end());
		int n = citations.size();
		for (int i = 0; i < n; i++) {
			if (citations[i] >= n - i) return n - i;
		}
		return 0;
	}
};

解法4 桶排序

这个方法比较取巧，代码如下：

class Solution {
public:
	int hIndex(vector<int>& citations) {
		vector<int> record(citations.size() + 1, 0);
		int n = citations.size();
		for (int citation : citations) {
			int id = min(citation, n);
			record[id]++;
		}
		int count = 0;
		int res;
		for (int i = 0; i <= n; i++) {
			if (n - count >= i) res = i;
			count += record[i];
		}
		return res;
	}
};