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.

Code

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;
    }
}

解题思路

• 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.