275. H-Index II (1)
Given an array of citations sorted in ascending order (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 = [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, her h-index is 3.
Note:
If there are several possible values for h, the maximum one is taken as the h-index.
Follow up:
This is a follow up problem to H-Index, where
citations
is now guaranteed to be sorted in ascending order.Could you solve it in logarithmic time complexity?
Solution
透過binary search去解決找出citation[i] >= len - i
public static int hIndex2(int[] citations){
Arrays.sort(citations);
for(int i = 0; i < citations.length; i++){
if(citations[i] >= (citations.length - i)){
return citations.length - i;
}
}
return 0;
}
將問題簡化成用有該citation以上的paper數當作index
ex:
citations [0 1 3 5 6] 原index 0 1 2 3 4 new index 5 4 3 2 1
然後從頭開始找當第一個new index >= citation時,new index就是那個 h (binary search)
要記住iteration結束後,left一定會指向造成h-index的位置,可以用len - left取得H value
public int hIndex(int[] citations) {
int len = citations.length;
int paperWithMoreCitationCount;
int left = 0;
int right = paperCount-1;
int mid;
while(right >= left){
mid = (left + right) /2 ;
paperWithMoreCitationCount = len - mid;
if(paperWithMoreCitationCount == citations[mid]){
return paperWithMoreCitationCount;
}
else if(paperWithMoreCitationCount > citations[mid]) {
//mid有可能是對的,但可以往右看看是否有更高的
left = mid+1;
}
else{
right = mid-1;
}
}
return paperCount - left;
//end the loop, the left will point the index that create h-index
}
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