[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;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/275

 

类似题目:

H-Index

 

参考资料:

https://leetcode.com/problems/h-index-ii/

https://leetcode.com/problems/h-index-ii/discuss/71063/Standard-binary-search

 

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posted @ 2015-09-05 05:45  Grandyang  阅读(12344)  评论(7编辑  收藏  举报
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