Leetcode 274/275 H-index Java__java

H-index I

Given an array of citations (each citation was a non-negative integer) of a researcher, write a function to compute the RES Earcher ' s h-index.

According to the definition of H-index on Wikipedia: "A scientist has index H. h of his/her N papers have at least h CIT Ations 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 into total and each of them had rece Ived 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 further 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.

This topic, prompted by the misleading, sorted, AP program is as follows:

    public int Hindex (int[] citations) {
        arrays.sort (citations);
        int rst=0;
        for (int i = citations.length-1 i >= 0; i--) {
            if (citations[i) >= citations.length-i)
                rst= CITATIONS.L ength-i;
        }
        return rst;
    }

However, in fact, there are other solutions to this problem, using space to change time, you can use O (n) time complexity to solve this issue. The procedure is as follows:

public int Hindex (int[] citations) {
    int len = citations.length;
    Int[] Count = new Int[len + 1];

    for (int c:citations)
        if (C > Len) 
            count[len]++;
        else 
            count[c]++;


    int total = 0;
    for (int i = len; I >= 0; i--) {Total
        = count[i];
        If (total >= i) return
            i;
    }

    return 0;
}

H-index II

Follow up to H-index:what if the citations array is sorted into ascending order? Could you optimize your algorithm?

Compared to the H-index I, the given array is already sorted.

So, direct two-point search can be.

public class Solution {public
    int hindex (int[] citations) {
        if (citations = null | | citations.length = 0) return 0;
        int L = 0, r = citations.length;
        int n = citations.length;
        while (L < r) {
            int mid = L + (r-l)/2;
            if (citations[mid] = = N-mid) return n-mid;
            if (Citations[mid] < Citations.length-mid) L = mid + 1;
            else R = Mid;
        }
        Return n-l
    }
}