274. H-Index

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题目内容

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return compute the researcher’s h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

Example 1:

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, their h-index is 3.
Example 2:

Input: citations = [1,3,1]
Output: 1

Constraints:

n == citations.length
1 <= n <= 5000
0 <= citations[i] <= 1000

方法

有两种方法解题排序和 [[计数排序]]

排序

时间复杂度和空间复杂度是排序的时间复杂度

class Solution {  
	fun hIndex(citations: IntArray): Int {  
		citations.sort()  
		var h = 0  
		var i = citations.size - 1;  
		while (i >= 0 && citations[i] > h) {  
			h++;  
			i--;  
		}  
		return h;  
    }  
}

计数排序

时间复杂度和空间复杂度都为O(N)

对每个可能的 h 值, 记录对应的论文数量

class Solution {  
	 fun hIndex(citations: IntArray): Int {  
		val n = citations.size  
		val counter = IntArray(n + 1)  

		counter.forEachIndexed { index, value ->  
			if (index >= n) {  
				return@forEachIndexed  
			}  
			if (citations[index] >= n) {
				counter[n]++;  
			} else {  
				counter[citations[index]]++;  
			}  
		}  

		var tot = 0;  

		var i = n  
		while (i >= 0) {  
			tot += counter[i];  
			if (tot >= i) {  
				return i  
			}  
			i--  
		}  
		return 0  
	 }  
}

参考

官方题解