LeetCode 274. H-Index

原题链接在这里：https://leetcode.com/problems/h-index/

题目：

Given an array of citations (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 = [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, her h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

题解：

两种方法，第一种方法是sorting, after sorting, could guess and use binary search.

Time Complexity O(nlogn). Space O(1).

AC Java:

class Solution {
    public int hIndex(int[] citations) {
        if(citations == null || citations.length == 0){
            return 0;
        }
        
        Arrays.sort(citations);
        int n = citations.length;
        int l = 0;
        int r = n - 1;
        while(l <= r){
            int mid = l + (r - l) / 2;
            if(citations[mid] == n - mid){
                return citations[mid];
            }else if(citations[mid] < n - mid){
                l = mid + 1;
            }else{
                r = mid - 1;
            }
        }
        
        return n - l;
    }
}

第二种方法是用空间换时间. count[i] 存储 被引用了i遍的文章有几个，最后total的含义就是总共有几篇文章引用次数大于等于i. 引用次数大于等于 len的都算在 count[len]里面.

Time Complexity is O(n), Space O(n).

AC Java:

public class Solution {
    public int hIndex(int[] citations) {
        /*
        //Method 1
        if(citations == null || citations.length == 0){
            return 0;
        }
        Arrays.sort(citations);
        int res = 0;
        for(int i = citations.length-1; i>=0; i--){
            if(citations[i] >= citations.length-i){
                res = citations.length-i;
            }
        }
        return res;
        */
        
        //Method 2
        if(citations == null || citations.length == 0){
            return 0;
        }
        int len = citations.length;
        //count 数组储存了被引用 i 遍的文章有几个，i是count的index
        int [] count = new int[len+1];
        for(int i = citations.length - 1; i>=0; i--){
            if(citations[i]>=citations.length){
                count[len]++;
            }else{
                count[citations[i]]++;
            }
        }
        //total 计数总共有几篇文章引用大于等于i次，i是count 的index
        int total = 0;
        for(int i = len; i>=0; i--){
            total += count[i];
            if(total>=i){ //有三篇文章引用大于等于三次
                return i;
            }
        }
        return 0;
    }
}

 跟上H-Index II.