274. H-Index(难)

一、

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.

一名科研人员的h指数是指他至多有h篇论文分别被引用了至少h次。

用hash[i]表示引用i次的文章数。如果i>n，用hash[n]计数。

从后往前遍历hash数组，sum累计hash[i]，找到第一个sum>=i的i返回。

class Solution {public:int hIndex(vector<int>& citations) {int n = citations.size();vector<int> hash(n+1,0);for (int i = 0; i < n; i++){if (citations[i]>=n){hash[n] += 1;}else{hash[citations[i]] += 1;}}int sum = 0;for (int i = n; i >= 0; i--){sum += hash[i];if (sum >= i){return i;}}return 0;}};

二、

275. H-Index II

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

注意最后的返回值

class Solution {public:int hIndex(vector<int>& citations) {int left = 0, right = citations.size() - 1;while (left <= right){int mid = left + (right - left) / 2;if (citations.size() - mid == citations[mid]){return citations.size()-mid;}else if (citations[mid] > citations.size() - mid){right = mid - 1;}else{left = mid + 1;}}return citations.size()-left;//}};