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Array
Counting Sort
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274. H-Index

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, 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.

 

Example 1:

Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 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 = [1,3,1]
Output: 1

 

Constraints:

  • n == citations.length
  • 1 <= n <= 5000
  • 0 <= citations[i] <= 1000

Solutions

Solution 1: Sorting

We can sort the array citations in descending order. Then we enumerate the value $h$ from large to small, if there is an $h$ value satisfying $citations[h-1] \geq h$, it means that there are at least $h$ papers that have been cited at least $h$ times, just return $h$ directly. If we cannot find such an $h$ value, it means that all the papers have not been cited, return $0$.

Time complexity $O(n \times \log n)$, space complexity $O(\log n)$. Here $n$ is the length of the array citations.

Python3

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        citations.sort(reverse=True)
        for h in range(len(citations), 0, -1):
            if citations[h - 1] >= h:
                return h
        return 0

Java

class Solution {
    public int hIndex(int[] citations) {
        Arrays.sort(citations);
        int n = citations.length;
        for (int h = n; h > 0; --h) {
            if (citations[n - h] >= h) {
                return h;
            }
        }
        return 0;
    }
}

C++

class Solution {
public:
    int hIndex(vector<int>& citations) {
        sort(citations.rbegin(), citations.rend());
        for (int h = citations.size(); h; --h) {
            if (citations[h - 1] >= h) {
                return h;
            }
        }
        return 0;
    }
};

Go

func hIndex(citations []int) int {
	sort.Ints(citations)
	n := len(citations)
	for h := n; h > 0; h-- {
		if citations[n-h] >= h {
			return h
		}
	}
	return 0
}

TypeScript

function hIndex(citations: number[]): number {
    citations.sort((a, b) => b - a);
    for (let h = citations.length; h; --h) {
        if (citations[h - 1] >= h) {
            return h;
        }
    }
    return 0;
}

Rust

impl Solution {
    #[allow(dead_code)]
    pub fn h_index(citations: Vec<i32>) -> i32 {
        let mut citations = citations;
        citations.sort_by(|&lhs, &rhs| rhs.cmp(&lhs));

        let n = citations.len();

        for i in (1..=n).rev() {
            if citations[i - 1] >= (i as i32) {
                return i as i32;
            }
        }

        0
    }
}

Solution 2: Counting + Sum

We can use an array $cnt$ of length $n+1$, where $cnt[i]$ represents the number of papers with the reference count of $i$. We traverse the array citations and treat the papers with the reference count greater than $n$ as papers with a reference count of $n$. Then we use the reference count as the index and add $1$ to the corresponding element of $cnt$ for each paper. In this way, we have counted the number of papers for each reference count.

Then we enumerate the value $h$ from large to small, and add the element value of $cnt$ with the index of $h$ to the variable $s$, where $s$ represents the number of papers with a reference count greater than or equal to $h$. If $s \geq h$, it means that at least $h$ papers have been cited at least $h$ times, just return $h$ directly.

Time complexity $O(n)$, space complexity $O(n)$. Here $n$ is the length of the array citations.

Python3

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)
        cnt = [0] * (n + 1)
        for x in citations:
            cnt[min(x, n)] += 1
        s = 0
        for h in range(n, -1, -1):
            s += cnt[h]
            if s >= h:
                return h

Java

class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int[] cnt = new int[n + 1];
        for (int x : citations) {
            ++cnt[Math.min(x, n)];
        }
        for (int h = n, s = 0;; --h) {
            s += cnt[h];
            if (s >= h) {
                return h;
            }
        }
    }
}

C++

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        int cnt[n + 1];
        memset(cnt, 0, sizeof(cnt));
        for (int x : citations) {
            ++cnt[min(x, n)];
        }
        for (int h = n, s = 0;; --h) {
            s += cnt[h];
            if (s >= h) {
                return h;
            }
        }
    }
};

Go

func hIndex(citations []int) int {
	n := len(citations)
	cnt := make([]int, n+1)
	for _, x := range citations {
		cnt[min(x, n)]++
	}
	for h, s := n, 0; ; h-- {
		s += cnt[h]
		if s >= h {
			return h
		}
	}
}

TypeScript

function hIndex(citations: number[]): number {
    const n: number = citations.length;
    const cnt: number[] = new Array(n + 1).fill(0);
    for (const x of citations) {
        ++cnt[Math.min(x, n)];
    }
    for (let h = n, s = 0; ; --h) {
        s += cnt[h];
        if (s >= h) {
            return h;
        }
    }
}

Solution 3: Binary Search

We notice that if there is a $h$ value that satisfies at least $h$ papers are cited at least $h$ times, then for any $h'&lt;h$, at least $h'$ papers are cited at least $h'$ times. Therefore, we can use the binary search method to find the largest $h$ such that at least $h$ papers are cited at least $h$ times.

We define the left boundary of binary search $l=0$ and the right boundary $r=n$. Each time we take $mid = \lfloor \frac{l + r + 1}{2} \rfloor$, where $\lfloor x \rfloor$ represents floor $x$. Then we count the number of elements in array citations that are greater than or equal to $mid$, and denote it as $s$. If $s \geq mid$, it means that at least $mid$ papers are cited at least $mid$ times. In this case, we change the left boundary $l$ to $mid$. Otherwise, we change the right boundary $r$ to $mid-1$. When the left boundary $l$ is equal to the right boundary $r$, we find the largest $h$ value, which is $l$ or $r$.

Time complexity $O(n \times \log n)$, where $n$ is the length of array citations. Space complexity $O(1)$.

Python3

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        l, r = 0, len(citations)
        while l < r:
            mid = (l + r + 1) >> 1
            if sum(x >= mid for x in citations) >= mid:
                l = mid
            else:
                r = mid - 1
        return l

Java

class Solution {
    public int hIndex(int[] citations) {
        int l = 0, r = citations.length;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            int s = 0;
            for (int x : citations) {
                if (x >= mid) {
                    ++s;
                }
            }
            if (s >= mid) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
}

C++

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int l = 0, r = citations.size();
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            int s = 0;
            for (int x : citations) {
                if (x >= mid) {
                    ++s;
                }
            }
            if (s >= mid) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
};

Go

func hIndex(citations []int) int {
	l, r := 0, len(citations)
	for l < r {
		mid := (l + r + 1) >> 1
		s := 0
		for _, x := range citations {
			if x >= mid {
				s++
			}
		}
		if s >= mid {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

TypeScript

function hIndex(citations: number[]): number {
    let l = 0;
    let r = citations.length;
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        let s = 0;
        for (const x of citations) {
            if (x >= mid) {
                ++s;
            }
        }
        if (s >= mid) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}