Question

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.

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

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;
  }
}

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;
  }
};

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
}

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;
}

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.

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

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;
      }
    }
  }
}

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;
      }
    }
  }
};

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
    }
  }
}

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 $h'<h$, 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 $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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;
}