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 has5
papers in total and each of them had
received 0, 1, 3, 5, 6
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.
Follow up:
- This is a follow up problem to H-Index, where
citations
is now guaranteed to be sorted in ascending order. - Could you solve it in logarithmic time complexity?
这题是之前那道 H-Index 的拓展,输入数组是有序的,让我们在 O(log n) 的时间内完成计算,看到这个时间复杂度,而且数组又是有序的,应该有很敏锐的意识应该用二分查找法,属于博主之前的总结帖 LeetCode Binary Search Summary 二分搜索法小结 中的第五类,目标值 target 会随着 mid 值的变化而变化,这里的 right 的初始值和 while 循环条件是否加等号是需要注意的问题,一般来说,博主的习惯是把 right 初始化为数组的长度,然后循环条件中不加等号,但是这种 right 的初始化对于这种目标值不固定的情况下不好使,需要初始化为长度减1(目前博主还没有遇到反例,有的话请务必告知博主)。那么此时循环条件中是否要加等号,这个其实很玄学,在 Find Peak Element 中,right 也是初始化为数组长度减1,但是循环条件却不能加等号。这道题却一定需要加等号,否则会跪在 [0] 这个 test case,有些时候固有的规律并不好使,可能只能代一些 corner case 来进行检验,比如 [], [0], [1,2] 这种最简便的例子。
基于上面的分析,我们最先初始化 left 和 right 为0和 len-1,然后取中间值 mid,比较 citations[mid] 和 len-mid 做比较,如果前者大,则 right 移到 mid 之前,反之 right 移到 mid 之后,循环条件是 left<=right,最后返回 len-left 即可,参见代码如下:
class Solution {
public:
int hIndex(vector<int>& citations) {
int len = citations.size(), left = , right = len - ;
while (left <= right) {
int mid = 0.5 * (left + right);
if (citations[mid] == len - mid) return len - mid;
else if (citations[mid] > len - mid) right = mid - ;
else left = mid + ;
}
return len - left;
}
};
类似题目:
参考资料:
https://leetcode.com/problems/h-index-ii/
https://leetcode.com/problems/h-index-ii/discuss/71063/Standard-binary-search