【leetcode】 H-Index II

1、题目描述

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?

2、问题描述：

• 求一个人的h因子，就是一个人的总共的文章为n，那么，h就是这n篇文章被引用次数超过h的最少有h篇。

3、问题关键：

• 有序数组，二分法。
• 第mid篇被引用的次数nums[mid],则 总共有（n - mid）篇因子超过nums[mid].
• 找到最小的mid。迭代条件，if(n - mid <= nums[mid]) r = mid;(左半区间）。

4、C++代码：

class Solution {
public:
    int hIndex(vector<int>& nums) {
        if (nums.empty()) return 0;
        int n = nums.size();
        int l = 0, r = n - 1;
        while(l < r) {
            int mid = l + r >> 1;
            if (n - mid <= nums[mid]) r = mid;//因为序号是0开始的，所以n - mid，是超过的数量。
            else l = mid + 1;
        }
        if (n - l <= nums[l]) return n - l;//n - l就是超过的数量。
        return 0;
    }
};