leetcode 274 H-Index || 275 H-Index II

leetcode 274 H-Index

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.

Hint:

1. An easy approach is to sort the array first.
2. What are the possible values of h-index?
3. A faster approach is to use extra space.

题目地址： leetcode H-Index

题意：

给定一个数组代表论文被引用次数。要求计算h-index。

h-index定义如下：

一个学者发表的n篇论文中，他有h篇被引用至少h次，而其他的n-h篇被引用次数都不超过h。

思路：

显然，答案的范围为[0,n]，n为数组的长度。

方法一

从高到底排序，然后对排完序的数组扫描。

对于i，h-index为min(i,citations[i])

C++

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

Python

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        if not citations: return 0
        return max([min(i + 1, c) for i, c in enumerate(sorted(citations, reverse=True))])

也可以

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        return sum([i < c for i, c in enumerate(sorted(citations, reverse=True))])

 

如果不倒着排,则可以这么写：

n - i 是大于引用数citations[i]的个数。

对于i，h-index 为min(n - i,citations[i])

C++

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;
	}
};

Python

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        if not citations: return 0
        return max([min(len(citations) - i, c) for i, c in enumerate(sorted(citations))])

 

 

方法二

对引用数进行计数。这样就不需要排序啦~

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        if not citations: return 0
        n = len(citations)
        cnt = [0] * (n + 1)
        for c in citations:
            if c < n:
                cnt[c] += 1
            else:
                cnt[n] += 1
        t = 0
        for i in xrange(n, -1, -1):
            t += cnt[i]
            if t >= i:
                return i
        return 0

 

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leetcode 275 H-Index II

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?

Hint:

• Expected runtime complexity is in O(log n) and the input is sorted.

题目地址： leetcode H-Index II

题意：

就是要求时间复杂度为O(log n) 的H-Index,数组已经升序排好

思路：

二分，枚举答案[0,n]

class Solution(object):
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        L, R, n = 0, len(citations), len(citations)

        while L < R:
            mid = (L + R) >> 1
            if n - mid <= citations[mid]:
                R = mid
            else:
                L = mid + 1

        return n - L