274. H-Index

题目：

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:

1. An easy approach is to sort the array first.
2. What are the possible values of h-index?
3. A faster approach is to use extra space.

链接： http://leetcode.com/problems/h-index/

题解：

求H-Index，找在数组中有k个元素大于等于k。排序一下再进行计算就比较容易些了。 也可以用index-couting，但这样就需要一个额外的数组。

Time Complexity - O(nlogn)， Space Complexity - O(1)。

public class Solution {
    public int hIndex(int[] citations) {
        if(citations == null || citations.length == 0) {
            return 0;
        }
        int len = citations.length;
        for(int i = 0; i < len; i++) {
            if(citations[i] >= len - i) {    // check if we have len - i elements larger than len - i
                return len - i;
            }
        }
        return 0;
    }
}

 

Index counting:

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public int hIndex(int[] citations) {
        if(citations == null || citations.length == 0) {
            return 0;
        }
        int len = citations.length;
        int[] count = new int[len + 1];
        for(int num : citations) {
            if(num > len) {
                count[len]++;
            } else {
                count[num]++;
            }
        }
        
        int sum = 0;
        for(int i = len; i >= 0; i--) {
            sum += count[i];
            if(sum >= i) {
                return i;
            }
        }
        
        return 0;
    }
}

 

Reference:

https://leetcode.com/discuss/55958/my-easy-solution

https://leetcode.com/discuss/56041/a-clean-o-n-solution-in-java

https://leetcode.com/discuss/55952/my-o-n-time-solution-use-java

https://leetcode.com/discuss/55950/1-line-ruby-5-lines-c-6-lines-java

https://leetcode.com/discuss/66656/java-o-n-time-with-easy-explanation