标签:index other public 数组 查找 ant rem value where
Given an array of citations sorted in ascending order (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 = [0,1,3,5,6]
Output: 3
Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had
received 0, 1, 3, 5, 6 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, her h-index is 3.
Note:
If there are several possible values for h, the maximum one is taken as the h-index.
Follow up:
citations
is now guaranteed to be sorted in ascending order.设一个研究者共有n篇论文,如果其中有h篇,这h篇中每一篇都被引用过至少h次,而剩余的n-h篇中每一篇被引用的次数都不超过h,则称h为这个研究者的h指数。给定一个研究者n篇论文引用次数的升序数组,求该研究者h指数的最大值。
与 0274. H-Index 相比,已经将数组按照升序排列,这样反而更简单。直接在 0274 的基础上使用二分查找进行改进即可。
class Solution {
public int hIndex(int[] citations) {
int left = 0, right = citations.length - 1;
while (left < right) {
int mid = (right - left) / 2 + left;
if (citations[mid] >= citations.length - mid) {
right = mid;
} else {
left = mid + 1;
}
}
return left == right && citations[left] >= citations.length - left ? citations.length - left : 0;
}
}
标签:index other public 数组 查找 ant rem value where
原文地址:https://www.cnblogs.com/mapoos/p/13190991.html