【LeetCode】274. H-Index 解题报告（Python）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址: https://leetcode.com/problems/h-index/description/

题目描述：

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.

题目大意

计算某个研究人员的影响因子。影响因子的计算方式是有h篇影响力至少为h的论文。影响因子是衡量作者生产力和影响力的方式，判断了他又多少篇影响力很大的论文。

解题方法

方法一：排序＋遍历

刚开始读不懂题，现在解释一下样例：[3,0,6,1,5]，当h=0时表示至少有0篇影响力为0的论文；当h=1时表示至少有1篇影响力为1的论文；当h=3时表示至少有3篇影响力为3的论文；当h=5时表示至少有5篇影响力为5的论文；当h=6时表示至少有6篇影响力为6的论文.显然符合要求的是只要有3篇影响力为3的论文。

有多少篇影响力大于x的论文怎么求呢？显然可以使用排序的方法，计算一下排序之后索引x后面有多少篇论文即可。我们求的结果就是求影响力x和不小于该影响力的论文个数的最小值，然后再求这个最小值的最大值。

时间复杂度是O(NlogN + N)，空间复杂度是O(N)。

class Solution(object):

    def hIndex(self, citations):

        """

        :type citations: List[int]

        :rtype: int

        """

        N = len(citations)

        citations.sort()

        h = 0

        for i, c in enumerate(citations):

            h = max(h, min(N - i, c))

        return h

方法二：排序+二分

这个题是275. H-Index II打乱了顺序的版本，也可以使用先排序，再二分的方法。这个做法打败了100%的提交。

class Solution(object):

    def hIndex(self, citations):

        """

        :type citations: List[int]

        :rtype: int

        """

        N = len(citations)

        citations.sort()

        l, r = 0, N - 1

        H = 0

        while l <= r:

            mid = l + (r - l) / 2

            H = max(H, min(citations[mid], N - mid))

            if citations[mid] < N - mid:

                l = mid + 1

            else:

                r = mid - 1

        return H

时间复杂度是O(NlogN + logN)，空间复杂度是O(N)。

参考资料：

https://blog.csdn.net/happyaaaaaaaaaaa/article/details/51593843

日期

2018 年 10 月 6 日 —— 努力看书

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