每个词的出现次数 [英] Number of occurences of each term

问题描述

我得到了一个数组 a [n] [2] ，其中 n 可以为<$ c最多$ c> 10 ^ 5 有n个科目和n个学生。全部编号为1，2，...，n。

I'm given an array a[n][2], where n can be 10^5 at max. There are n subjects and n students. All numbered 1, 2, ..., n.

a [i] [0] 和 a [i] [1] （ 1< = i< = n ）表示在第i个主题中，从 a [i] [0] 到 a [i] [1] 的所有学生都通过了，其余的则没有没错我应该找到每个学生通过的科目数量。

a[i][0] and a[i][1] (1 <= i <= n) denote that in i-th subject, all students from a[i][0] to a[i][1] passed, while the rest didn't. I'm supposed to find the number of subjects each student passed in.

例如，

n=5 and a = [ [1,3], [1,3], [1,5], [1,5], [1,5] ] 

应该给出输出

[5, 5, 5, 3, 3]

（2）

n = 10, a = [ [1,10], [1,10], [1,10], [3,5], [1,10], ..., [1,10] ]

预期答案应为

[ 9, 9, 10, 10, 10, 9, 9, 9, 9, 9]

推荐答案

不是很了解您的代码，这是另一种方法。
这类似于间隔问题。让我们举个例子：

Didn't quite understand your code, here is an alternative approach. This is similar to an interval problem. Let's take your example:

• n = 5


• a = [[1,3]， [1,3]，[1,5]，[1,5]，[1,5]]

我们首先大小为no的数组。的主题数+ 1（为了简化计算）。

We first make an array of size as no. of subjects + 1 (for the sake of simplicity in computation).

1   2   3   4   5   6

• 现在，让我们一个一个地讲一下。每当遇到间隔时，我们将 +1 添加到起点，并将 -1 添加到终点+ 1 ^th 位置。


• Now, let's go through intervals one by one. Whenever we come across an interval, we add +1 to the starting point and -1 to the ending point + 1^th location.

数组表示形式：（经过上述整个过程） 。

Array representation:(after the entire above process).

 1          2          3          4         5         6

+1                               -1                           [1,3]
+1                               -1                           [1,3]
+1                                                   -1       [1,5]
+1                                                   -1       [1,5]
+1                                                   -1       [1,5]







• 所有待处理就是从左到右开始求和得到每个学生的答案。上面的过程之所以有效，是因为我们在每个时间间隔的最后一个学生之后否定了（如-1一样），因为其他学生未通过该特定主题。因此，在求和时，我们一定会为每个学生得出正确的总数。



总结看起来像：

 1          2          3          4         5         6

+1                               -1                           [1,3]
+1                               -1                           [1,3]
+1                                                   -1       [1,5]
+1                                                   -1       [1,5]
+1                                                   -1       [1,5]

 5          5          5          3         3         0(not considered anyway) 

这篇关于每个词的出现次数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！