如何编写一个方法来查找范围数组中的所有交集？

发布于 2024-12-12 19:43:30 字数 332 浏览 0 评论 0原文

我有与此类似的数组

[
 [0, 10]**,
 [1, 3]**,
 [5, 11]**,
 [14, 20]**,
 [10, 11]**
]

** 表示包含数组中显示的开始和结束索引的对象

现在，交集是 [1, 3], [5,10], [10,11]

编写返回包含相交集的对象的方法的最佳方法是什么？（在我们进行过程中，可以将它们存储在一系列冲突的东西中）

我遇到的最大问题是如何做到这一点，以便每个对象都与彼此的对象进行比较？

有n！做到这一点的方法（我想，我对我的组合学有点生疏）

原文

I have array similar to this

[
 [0, 10]**,
 [1, 3]**,
 [5, 11]**,
 [14, 20]**,
 [10, 11]**
]

** Denotes an object containing the start and end indexes shown in the array

Now, the intersections are [1, 3], [5,10], [10,11]

What is the best way to write a method that returns the objects containing of the intersecting sets?
(Could just store them in an array of conflicted stuff as we go along)

The biggest issue I'm having is how do I do this such that each object is compared with eachother object?

there are n! ways to do this (i think, I'm a bit rusty on my combinatorics)

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梦里南柯 2024-12-19 19:43:30

按开始时间对间隔进行排序（或者按开始时间对索引数组进行排序，这样就不会丢失索引）。

之后，您可以执行一次检测所有交叉点/冲突。

Range array[N];

sort(array);
int i=0;
while(i<N){
    Range curr = array[i];  //invariant: curr doesn't intersect with anyone befor it
    i++;
    while(i < N && array[i].start <= curr.end){
        output that curr and array[i] intersect;
        if(array[i].end > curr.end){
            curr = array[i]
        }
        i++;
    }
}

Sort the intervals by start time (or instead sort an array of indices by start time so you don't lose the indices).

After this you can do a single pass detecting all intersections/conflicts.

Range array[N];

sort(array);
int i=0;
while(i<N){
    Range curr = array[i];  //invariant: curr doesn't intersect with anyone befor it
    i++;
    while(i < N && array[i].start <= curr.end){
        output that curr and array[i] intersect;
        if(array[i].end > curr.end){
            curr = array[i]
        }
        i++;
    }
}

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