【LeetCode从零单刷】H-index I & II

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 thanh 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.

解答：

一开始做的非常麻烦，遍历引用数（并没连续分布，有间隔值），因此还要计算 maximum、维护与 size 的关系。

遍历应针对连续值（此题中的 size 值是连续分布的）。其实看维基可以清楚知道算法：h-index (f) = 

class Solution {public:    int hIndex(vector<int>& citations) {        int size = citations.size();        if (size == 0)  return 0;                sort(citations.begin(), citations.end(), [](int a, int b){return a>b;});                vector<int> tmp;        for(int i = 0; i< size; i++)        {            tmp.push_back(citations[i]>(i+1)?(i+1):citations[i]);        }        sort(tmp.begin(), tmp.end(), [](int a, int b){return a>b;});                return tmp[0];    }};

H-index II 题目有着进一步要求，变为如下：

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?

Hint:

1. Expected runtime complexity is in O(log n) and the input is sorted.

解答：

看到 O（log n）应该自动想到二分搜索。而且这里是已经排好序的 citations 数组。注意处理数组为0，以及数组最大值小于等于0的情况。

class Solution {public:    int hIndex(vector<int>& citations) {        int size = citations.size();        if(size == 0 || citations[size - 1] <= 0)   return 0;                int left    = 0;        int right   = size - 1;        int mid;        while(left < right)        {            mid = (left + right) / 2;            if (citations[mid] < size - mid)                left = mid + 1;            else                right = mid;        }        return (size - right);    }};