[LeetCode ] H-Index

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最新推荐文章于 2024-01-03 13:28:28 发布

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本文介绍了一种计算研究者h指数的方法，通过两种算法实现：一是利用计数和后缀和技巧，时间复杂度为O(N)；二是通过排序和逆向遍历，时间复杂度为O(NlogN)。h指数反映了研究者发表论文的质量和数量，是衡量学术影响力的重要指标。

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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 leasth citations each, and the other N − h papers have no more than h citations each."

Example:

Input: citations = [3,0,6,1,5]Output: 3 
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3 citations each, her h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

题意：给出一个n个元素的数组，找出一个h，这个数组有h个数大于等于h，剩下h个数小于等于h，输出最大的h。

思路：

解法一：再定义一个数组，记录每个元素出现了多少次，再对这个数组求后缀和，然后从后向前遍历，如果后缀和大于等于i，停止遍历。时间复杂度O（N）。

C代码：

int hIndex(int* citations, int citationsSize) {
    int ans = 0,i;
    int pre[1005];
    memset(pre,0,sizeof(pre));
    for(i = 0; i < citationsSize; i++) {
        pre[citations[i]]++;
    }
    for(i = 1000; i >= 0; i--) {
        pre[i] = pre[i + 1] + pre[i];
        if(pre[i] >= i) {
            ans = i;
            break;
        }
    }
    return ans;
}

解法二：先对数组排序，从后向前遍历，如果此时枚举到的数大于等于已经枚举的数的个数，停止遍历。时间复杂度O（NlogN）。

Java代码：

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