H-Index I & II
H-Index I
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：
先排序，然后看倒数第n个的paper它的citation是否超过n.
public class Solution {
public int hIndex(int[] citations) {
// 排序
Arrays.sort(citations);
int h = ;
for(int i = ; i <= citations.length; i++){
// 最后i个paper的最小index是否大于i
if (citations[citations.length - i] >= i) {
h = Math.max(h, i);
}
}
return h;
}
}
H-Index II
Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?
public class Solution {
public int hIndex(int[] citations) {
int start = ;
int end = citations.length - ;
while (start <= end) {
int mid = (start + end) / ;
if (citations[mid] >= citations.length - mid) {
end = mid - ;
} else {
start = mid + ;
}
}
return citations.length - start;
}
}
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