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[Leetcode]275. H-Index II

题目描述

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.”

例子

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 hadreceived 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.

FollowUp

  • 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?

解题思路

跟上一道 H-Index 题类似,只要找到符合 H-Index 的定义的话,这道题不算很难,由于这道题已经排好序了,可以参考上一道题排序的做法,再通过二分查找提高时间复杂度,我的代码如下:

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class Solution {
public:
int hIndex(vector<int>& citations) {

int left = 0, right = citations.size() - 1;
int result = 0;

while (left <= right) {
int mid = (left + right) / 2;

/// candidate h-index
int h = citations.size() - mid;

if (citations[mid] >= h) {
// current h is a valid h-index, we need to choose the largest one
result = max(result, h);
right = mid - 1;
} else {
left = mid + 1;
}
}

return result;

}
};
  • 时间复杂度: O(logN)
  • 空间复杂度: O(1)