[LeetCode] 275. H-Index II

轉自LeetCode

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.

<Soluion>

H-Index 的衍生題

解題想法：

• 這次給的 input 已經是排序好的，所以可以直接拿來利用

• 從題目來看，其實就是要找到一個一刀兩斷的 index，讓在這個點之上的所有值，都大於等於這個 index，在這個點之下的值，都小於等於這個 index

• 綜合上述兩點，可以使用 binary search 來找，配合題意檢查和回傳對的值即可

code 如下

Java

	class Solution {
	public int hIndex(int[] citations) {
	int left = 0, right = citations.length;
	int mid, diff;
	while(left < right) {
	mid = left + (right - left) / 2;
	diff = citations.length - mid;
	if(diff == citations[mid]) {
	return diff;
	} else if(diff > citations[mid]){
	left = mid+1;
	} else {
	right = mid;
	}
	}
	return citations.length - left;
	}
	}

view raw hIndex_2.java hosted with ❤ by GitHub

kotlin

	class Solution {
	fun hIndex(citations: IntArray): Int {
	var left = 0
	var right = citations.size
	while(left < right) {
	val mid = left + (right - left) / 2
	val diff = citations.size - mid
	when {
	diff == citations[mid] -> return diff
	citations[mid] < diff -> left = mid+1
	else -> right = mid
	}
	}
	return citations.size - left
	}
	}

view raw h_index_II.kt hosted with ❤ by GitHub