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package LeetCode_HashTable;/*** 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* [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.* （3,0,6,1,5]表示研究人员共有5篇论文，每篇论文分别获得3次，0次，6次，1次5次引用。由于研究人员有3篇论文，每篇论文至少引用3篇，* 其余两个每个引用次数不超过3次，她的h指数为3。）* 题目大意：* 给定研究者的引用数组（每个引用是非负整数），编写一个函数来计算研究者的h指数。* 根据维基百科上h-index的定义：“如果他/她的N篇论文中至少有h引文，那么科学家就有索引h，其他的N-h论文每篇都不超过引 数。”* 解题思路：* 根据给出的数组citations创建一个新的数组array，使新数组的索引代表论文被引用的次数，数组值代表引用次数出现的次数。* 然后从新数组的末尾开始向数组的头部开始遍历，在每次遍历时，将数组值累加求和sum,若出现所求和大于数组索引，* 则此时的数组索引即为所求的h指数。*/
public class HIndex_274_1022 {public int HIndex(int[] citations) {if (citations == null || citations.length == 0) {return 0;}int length = citations.length;int[] array = new int[length + 1];for (int i = 0; i < length; i++) {if (citations[i] >= length) {array[length] += 1;} else array[citations[i]] += 1;}int result ; int num = 0;for (result = length; result >= 0; result--) {num += array[result];if (num >= result) {return result;}}return 0;}public static void main(String[] args) {int[] array = {3, 0, 6, 1, 5};HIndex_274_1022 test = new HIndex_274_1022();int result = test.HIndex(array);System.out.println(result);}
}
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