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."
Example:
Input:citations = [3,0,6,1,5]
Output: 3 Explanation:[3,0,6,1,5]
means the researcher has5
papers in total and each of them had received3, 0, 6, 1, 5
citations respectively. Since the researcher has3
papers with at least3
citations each and the remaining two with no more than3
citations each, her h-index is3
.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
public class Solution { public int hIndex(int[] citations) { ArrayList<Integer> list = new ArrayList(); for(int i: citations){ list.add(i); } Collections.sort(list); Collections.reverse(list); for (int i = 0; i < citations.length; ++i) { if (i + 1 == list.get(i)) return i+1; if (i + 1 > list.get(i)) return i; //[0]的情况,很傻逼 } return citations.length; } }
这个方法反而慢的一比
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; //[0]的情况,傻逼 } 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; } } }
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O(nlog,空间复杂度O(1)。