LeetCode 274/275 H-Index Java

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.

这道题目，受提示的误导，排序了，AP程序如下：

public int hIndex(int[] citations) {
Arrays.sort(citations);
int rst=0;
for (int i = citations.length - 1; i >= 0; i--) {
if (citations[i] >= citations.length - i)
rst= citations.length - i;
}
return rst;
}

但是，其实这道题目还有其他解法，用空间换时间，可以用o（n）的时间复杂度解决这个问题。程序如下：

public int hIndex(int[] citations) {
int len = citations.length;
int[] count = new int[len + 1];

for (int c: citations)
if (c > len)
count[len]++;
else
count[c]++;

int total = 0;
for (int i = len; i >= 0; i--) {
total += count[i];
if (total >= i)
return i;
}

return 0;
}

H-Index II

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

和H-Index I 相比，给定的数组已经排好序了。

所以，直接二分查找即可。

public class Solution {
public int hIndex(int[] citations) {
if(citations == null || citations.length == 0) return 0;
int l = 0, r = citations.length;
int n = citations.length;
while(l < r){
int mid = l + (r - l) / 2;
if(citations[mid] == n - mid) return n - mid;
if(citations[mid] < citations.length - mid) l = mid + 1;
else r = mid;
}
return n - l;
}
}