275. H-Index II

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? 275. H-Index II

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是LeetCode H-Index的延伸，如果citations是按升序排列的，怎样把算法优化到O(logn)。看到O(logn)，马上想到二分查找。 首先左右边界是0和n-1，判断中点的citations[m]和n-m，如果正好相等，则说明正好有n-m篇大于等于n-m引用的文章，所以H-index为n-m。如果citations[m]>n-m，说明引用很大，h-index还能往上涨，所以中心点需要左移，即r=m-1。反之则中心点右移l=m+1。
代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations)
    {
        int l = 0, n = citations.size(), r = n – 1;
        while (l <= r) {
            int m = l + (r – l) / 2;
            if (n – m == citations[m])
                return n – m;
            else if (citations[m] > n – m)
                r = m – 1;
            else
                l = m + 1;
        }
        return n – l;
    }
};

本代码提交AC，用时9MS。
我一开始的代码是，如果citations[m]<n-m，中心点确实右移了，即l=m+1。但是当citations[m]>n-m时，我没想到用r=m-1来左移中心点，误以为r一直都是n-1是固定的，所以这个时候使用了一个while循环来线性查找正确的m，理论上性能要比上面的纯二分差。
代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations)
    {
        int l = 0, r = citations.size() – 1, n = citations.size();
        if (n == 0)
            return 0;
        if (citations[0] >= n)
            return n;
        while (l < r) {
            int m = l + (r – l) / 2;
            if (n – m > citations[m])
                l = m + 1;
            else {
                while (m > 0 && n – m <= citations[m])
                    –m;
                return n – m – 1;
            }
        }
        if (citations[r] >= 1)
            return 1;
        else
            return 0;
    }
};

本代码提交AC，用时也是9MS。

二刷。更好理解的二分，因为是从小到大排序的，选定中点m后，中点及其右边共有n-m篇paper，如果paper数和citation相等，正好；如果citations[m]更大，则还可以增加paper数，所以r=m-1区间左移，右边合法的paper区间增大。因为如果合法的话，区间会一直左移，r会一直m-1，直到不满足要求，退出while循环，所以最后一个合法的下标r+1，则右边的paper数为n-(r+1)。

class Solution {
public:
	int hIndex(vector<int>& citations) {
		int n = citations.size();
		int l = 0, r = n - 1;
		while (l <= r) {
			int m = l + (r - l) / 2;
			int papers = n - m;
			if (citations[m] == papers) {
				return papers;
			}
			else if (citations[m] >= papers) {
				r = m - 1;
			}
			else {
				l = m + 1;
			}
		}
		return n - (r + 1);
	}
};