H-Index(II)

• H-Index
  • 题目描述
  • 解法
• H-Index II
  • 题目描述
  • 解法

H-Index

题目描述

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.

解法

代码如下:

public static int hIndex( int[] citations ) {
    Arrays.sort( citations );
    for( int i = 0; i < citations.length; ++i )
        if( citations[i] >= citations.length - i )
            return citations.length - i;
    return 0;
}

思考过程:

根据题意, 知道h-index与元素个数和数值大小有关. 所以可以先排序.

元素的个数可以根据数组下标确定, 数值大小根据元素值确定.

如果当前元素值大于等于元素的个数, 说明后面的元素也都大于等于元素的个数, 这就说明找到h-index

时空复杂度: 时间复杂度是O(nlogn), 空间复杂度是O(1)

H-Index II

题目描述

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?

Hint:

Expected runtime complexity is in O(logn) and the input is sorted.

解法

代码如下:

public static int hIndex( int[] citations ) {
    int low = 0, high = citations.length-1;
    while( low <= high ) {
        int mid = low + ( high - low ) / 2;
        if( citations.length - mid <= citations[mid] )
            high = mid - 1;
        else
            low = mid + 1;
    }
    return citations.length - low;
}

思考过程: 二分搜索.跟上面一样的思考方式.

每次找到范围内的中间值, 如果中间值大于等于元素个数, 为了获得更大的h-index, 向左边继续搜索.

时空复杂度: 时间复杂度是O(logn), 空间复杂度是O(1)