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.
class Solution {
public int hIndex(int[] citations) {
int n = citations.length;
int[] buckets = new int[n+1];
for(int c : citations) {
if(c >= n) {
buckets[n]++;
} else {
buckets[c]++;
}
}
int count = 0;
for(int i = n; i >= 0; i--) {
count += buckets[i];
if(count >= i) {
return i;
}
}
return 0;
}
}
- 使用buckets排序,新建一个len+1的数组;
- 遍历原数组,在新数组元素对应位加一,如果元素超出数组长度,在新数组的末尾加一;
- 从后向前遍历新数组,如果找到了元素相加大于等于索引为的元素,则返回索引值,否则返回0.