[LeetCode] 274. H-Index

轉自LeetCode

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 hof 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.

<Solution>

想法如下:

• 將 citation 做分類，分裝到不同的 bucket

• 為什麼選擇用 bucket 的方式分類，因為這邊只在乎在某個數以上的paper數量，和以下的 paper 數量，所以利用 bucket 分類，可以很快計算出以上和以下的數量

• 因為 paper 的數量，剛好等於 array 的長度，所以 bucket 的數量，設為 citations.length + 1，把比 paper 數量多的 citation 都放在同一個 bucket

• 最後，因為題目要求要找最大的 h-index，所以從最後面的 bucket 往前找。當在該 bucket 的數目，已經大於等於該 bucket 的 index時，就找到最大的 h-index 了

code 如下

Java(參考方法)

	class Solution {
	public int hIndex(int[] citations) {
	final int len = citations.length;
	final int[] buckets = new int[len+1];
	for(final int c : citations) {
	if(c >= len) {
	buckets[len]++;
	} else {
	buckets[c]++;
	}
	}
	int cnt = 0;
	for(int i = len; i >= 0; i--) {
	cnt += buckets[i];
	if(cnt >= i) {
	return i;
	}
	}
	return 0;
	}
	}

view raw hIndex.java hosted with ❤ by GitHub

Kotlin

	class Solution {
	fun hIndex(citations: IntArray): Int {
	val buckets = IntArray(citations.size+1) {0}
	for(c in citations) {
	if(c >= citations.size) {
	buckets[citations.size] += 1
	} else {
	buckets[c] += 1
	}
	}
	var sum = 0
	for(i in buckets.lastIndex downTo 0) {
	sum += buckets[i]
	if(sum >= i) {
	return i
	}
	}
	return 0
	}
	}

view raw h_index.kt hosted with ❤ by GitHub