[LeetCode] 275. H-Index II 求H指数之二

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in an ascending order, return compute the researcher's h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

You must write an algorithm that runs in logarithmic time.

Example 1:

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, their h-index is 3.

Example 2:

Input: citations = [1,2,100]
Output: 2

Constraints:

n == citations.length
1 <= n <= 105
0 <= citations[i] <= 1000
citations is sorted in ascending order.

这题是之前那道 H-Index 的拓展，输入数组是有序的，让我们在 O(log n) 的时间内完成计算，看到这个时间复杂度，而且数组又是有序的，应该有很敏锐的意识应该用二分查找法，属于博主之前的总结帖 LeetCode Binary Search Summary 二分搜索法小结中的第五类，目标值 target 会随着 mid 值的变化而变化，这里的 right 的初始值和 while 循环条件是否加等号是需要注意的问题，一般来说，博主的习惯是把 right 初始化为数组的长度，然后循环条件中不加等号，但是这种 right 的初始化对于这种目标值不固定的情况下不好使，需要初始化为长度减1（目前博主还没有遇到反例，有的话请务必告知博主）。那么此时循环条件中是否要加等号，这个其实很玄学，在 Find Peak Element 中，right 也是初始化为数组长度减1，但是循环条件却不能加等号。这道题却一定需要加等号，否则会跪在 [0] 这个 test case，有些时候固有的规律并不好使，可能只能代一些 corner case 来进行检验，比如 [], [0], [1,2] 这种最简便的例子。

基于上面的分析，我们最先初始化 left 和 right 为0和 len-1，然后取中间值 mid，比较 citations[mid] 和 len-mid 做比较，如果前者大，则 right 移到 mid 之前，反之 right 移到 mid 之后，循环条件是 left<=right，最后返回 len-left 即可，参见代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int len = citations.size(), left = 0, right = len - 1;
        while (left <= right) {
            int mid = 0.5 * (left + right);
            if (citations[mid] == len - mid) return len - mid;
            else if (citations[mid] > len - mid) right = mid - 1;
            else left = mid + 1;
        }
        return len - left;
    }
};