README_EN.md

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275. H-Index II

中文文档

Description

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in ascending order, return the researcher's h-index.

According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h papers that have each been cited at least h times.

You must write an algorithm that runs in logarithmic time.

 

Example 1:

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, their h-index is 3.

Example 2:

Input: citations = [1,2,100]
Output: 2

 

Constraints:

• n == citations.length
• 1 <= n <= 105
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

Solutions

Solution 1: Binary Search

We notice that if there are at least $x$ papers with citation counts greater than or equal to $x$, then for any $y \lt x$, its citation count must also be greater than or equal to $y$. This exhibits monotonicity.

Therefore, we use binary search to enumerate $h$ and obtain the maximum $h$ that satisfies the condition. Since we need to satisfy that $h$ papers are cited at least $h$ times, we have $citations[n - mid] \ge mid$.

The time complexity is $O(\log n)$, where $n$ is the length of the array $citations$. The space complexity is $O(1)$.

Python3

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)
        left, right = 0, n
        while left < right:
            mid = (left + right + 1) >> 1
            if citations[n - mid] >= mid:
                left = mid
            else:
                right = mid - 1
        return left

Java

class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (citations[mid] >= n - mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return n - left;
    }
}

C++

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (citations[n - mid] >= mid)
                left = mid;
            else
                right = mid - 1;
        }
        return left;
    }
};

Go

func hIndex(citations []int) int {
	n := len(citations)
	left, right := 0, n
	for left < right {
		mid := (left + right + 1) >> 1
		if citations[n-mid] >= mid {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return left
}

TypeScript

function hIndex(citations: number[]): number {
    const n = citations.length;
    let left = 0,
        right = n;
    while (left < right) {
        const mid = (left + right + 1) >> 1;
        if (citations[n - mid] >= mid) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
}

Rust

impl Solution {
    pub fn h_index(citations: Vec<i32>) -> i32 {
        let n = citations.len();
        let (mut left, mut right) = (0, n);
        while left < right {
            let mid = ((left + right + 1) >> 1) as usize;
            if citations[n - mid] >= (mid as i32) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        left as i32
    }
}

C#

public class Solution {
    public int HIndex(int[] citations) {
        int n = citations.Length;
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (citations[n - mid] >= mid) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}