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:
- An easy approach is to sort the array first.
- What are the possible values of h-index?
- A faster approach is to use extra space.
solution:
1. O(nlog(n)), sort array, use definition to find maximum index.
2. use extra space, O(n). counting sort
public class Solution {
public int hIndex(int[] citations) {
if(citations.length <=0) return 0;
int res = 0;
int len = citations.length;
int[] countarray = new int[len+1];
for(int i=0;i<len;i++) {
if(citations[i]>=len) {
countarray[len]++;
}else {
countarray[citations[i]]++;
}
}
if(countarray[len]>=len) return len;
for(int i=len-1;i>=0;i--){
countarray[i] = countarray[i] + countarray[i+1];
if(countarray[i]>=i) return i;
}
return res;
}
}

本文介绍了一种计算研究人员h指数的方法。h指数是指在N篇论文中,有h篇论文至少被引用了h次,而其余的N-h篇论文每篇被引用不多于h次。文章提供了两种解决方案:一种是对引用次数进行排序并根据定义找到最大索引;另一种是使用额外的空间进行计数排序。
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