274. H-Index 275. H-Index II

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."

For example, given citations = [3, 0, 6, 1, 5], which 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, his h-index is 3.

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

Hint:

1. An easy approach is to sort the array first.
2. What are the possible values of h-index?
3. A faster approach is to use extra space.

我是看了第二条hint才想到的。这属于结果在一个小范围内的题。

因为可能的h-index 最大也就是n

那么构造一个count 数列，如果citation次数大于n，count加到count[n]上，else 记录那个citation[i]（例如5）的count 次数。

想要知道最大count >= i， 从右边向左扫，count[i-1]+=count[i]

  public int hIndex(int[] citations) {  
     int n = citations.length;  
     int[] count = new int[n+1];  
     for(int i = 0; i < n; i++){  
       if(citations[i] >= n)  
         count[n]++;  
       else  
         count[citations[i]]++;  
     }  
     for(int i = n; i> 0; i--){  
       if(count[i] >= i) return i;  
       count[i-1] += count[i];  
     }  
     return 0;  
   }  

Follow up for H-Index: What if the citations array is sorted in ascending order? 

Could you optimize your algorithm?

解法：
citations[i]来说，这个位置的结果是Math.min(n-i,citations[i])
，因此就是找位置使得Math.min(n-i,citations[i])最大。
当citations[i]值比右边的个数大的时候，结果不可能在这个位置的右边，
即左边还有提高H-index的空间，因为count还有上升空间。
当citations[i]值比右边的个数小的时候，结果不可能在这个位置的左边。

看了hint才知道，因为是sorted array 可以binary search

   public int hIndex(int[] citations) {  
     int lo = 0, n =citations.length ,hi = n-1, mid =0,max = 0;  
     while(lo <= hi){  
       mid = lo + (hi - lo)/2;  
       if(citations[mid]==n-mid){  
         return n-mid;  
       }else if(citations[mid]>n-mid){  
         max = Math.max(max,n-mid);  
         hi = mid - 1;  
       }else{  
         max = Math.max(max,citations[mid]);  
         lo = mid + 1;    
       }  
     }  
     return max;  
   }