H-Index
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.
分析
H-Index的含义是，如果一个人发表的所有论文中，有h篇论文分别被引用了至少h次，那么他的H-Index就是h。
思路一：先从大到小排序，然后从前往后扫描，如果当前文章数（即当前下标+1）等于值本身，则返回当前文章数作为 h-index；如果当前文章数大于值本身，则返回当前文章数-1作为H-Index, 因为当前文章的引用数小于当前文章数，不能算在内。时间复杂度，空间复杂度O(1)。
思路二：跟思路一类似，不过排序算法换成了计数排序。有一个小技巧，因为H-Index最大不可能超过论文综述，所以我们只需要开一个长度为n+1的数组，如果某篇论文的引用数超过了n，就将其当做n。
代码1 全排序
// H-Index
// Time complexity: O(nlogn), Space complexity: O(1)
public class Solution {
public int hIndex(int[] citations) {
Arrays.sort(citations);
reverse(citations);
for (int i = 0; i < citations.length; ++i) {
if (i + 1 == citations[i]) return i+1;
if (i + 1 > citations[i]) return i;
}
return citations.length;
}
private static void reverse(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
final int tmp = nums[left];
nums[left] = nums[right];
nums[right] = tmp;
++left;
--right;
}
}
}代码2 计数排序
// H-Index
// Time complexity: O(n), Space complexity: O(n)
public class Solution {
public int hIndex(int[] citations) {
final int n = citations.length + 1;
final int[] histogram = new int[n+1];
for (int x : citations) {
++histogram[x > n ? n : x];
}
int sum = 0; // current number of papers
for (int i = n; i > 0; --i) {
sum += histogram[i];
if (sum >= i) {
return i;
}
}
return 0;
}
}