C++ 排序间隔结构以及删除重叠间隔

Question

这里的intervals和Line是一样的。尽管我为这两个目的找到了正确的解决方案:

1) 排序间隔；

2) 添加与现有区间重叠的区间时拒绝。

struct Line {
    int down;
    int up;

    Line(int down, int up) {
        this->down = down;
        this->up = up;
    }
};

int main() {
    auto cmp = [] (const Line& a, const Line& b) {//NOTE: this can even detect the intersection, as well as sort the interval!!
        if (a.up <= b.down) return true;
        else return false;
    };

    set<Line, decltype(cmp)> s(cmp);
    Line l1{2, 3};
    Line l2{3, 5};
    Line l3{1, 2}; //success for insertion
    Line l4{4, 6}; //fail for insertion b/c of insersection


    auto ret = s.insert(l2);
    cout << "insert 3->5 ";
    if (ret.second) cout << "success " << endl;
    else cout << "fail " << endl;

    ret = s.insert(l3);
    cout << "insert 1->2 ";
    if (ret.second) cout << "success " << endl;
    else cout << "fail " << endl;

    ret = s.insert(l1);
    cout << "insert 2->3 ";
    if (ret.second) cout << "success " << endl;
    else cout << "fail " << endl;

    ret = s.insert(l4);
    cout << "insert 4->6 ";
    if (ret.second) cout << "success " << endl;
    else cout << "fail " << endl;

    return 0;
}

Output:
insert 3->5 success
insert 1->2 success
insert 2->3 success
insert 4->6 fail
set finally contains: 1->2, 2->3, 3->5,

还有一点让我费解:

1) 为什么只是比较a.up <= b.down足以检测重叠间隔？

我们不应该做一些类似检测间隔重叠的标准方法吗？

max(a.down, b.down) < min(a.up, b.up)

2)第三步我不清楚

insert 3->5 success
insert 1->2 success //OK, b/c 2 <= 3, a.up <= b.down return true
insert 2->3 success //Why success? 3 > 1, i.e., a.up > b.down, return false; does it only compare with interval 3->5?

3) 为什么下面的代码不能去除重叠间隔？

    auto cmp_set = [](const Line& a, const Line& b) {
        if (max(a.down, b.down) < min(a.up, b.up)) return 0; //standard way to detect overlapping of intervals
        else { //sort the two intervals 
            if(a.up <= b.down) return -1;
            else return 1;
        }
    };

Answer 1

1) Why is simply comparing a.up <= b.down enough to detect overlapping intervals?

[a.down, a.up[重叠 [b.down, b.up[什么时候max(a.down, b.down) < min(a.up, b.up) 相当于a.down < b.up && b.down < a.up (请记住，我们还有来自区间定义的 a.down <= a.up && b.down <= b.up)。

我们还可以列出可能性:

• a.down < a.up < b.down < b.up (a < b)
• a.down < b.down < a.up < b.up路口
• a.down < b.down < b.up < a.up交点(全部包含)
• b.down < a.down < a.up < b.up交点(全部包含)
• b.down < a.down < b.up < a.up路口
• b.down < b.up < a.down < a.up (b < a)

The 3rd step is not clear to me

insert 3->5 success insert 1->2 success //OK, b/c 2 <= 3, a.up <= b.down return true insert 2->3 success //Why success? 3 > 1, i.e., a.up > b.down, return false; does it only compare with interval 3->5?

Line l1{2, 3};
Line l2{3, 5};
Line l3{1, 2};
Line l4{4, 6};

插入l1在 {l3, l2}

• l1 < l2所以l1在 l2 之前
• !(l1 < l3)所以l1不在 l3 之前
• l3 < l1所以l1在l3之后 -> {l3, l1, l2}

对于 l4在 {l3, l1, l2}我们有:

• !(l4 < l2) (类似于 l3 和 l1 )所以 l4不在 l2 之前
• !(l2 < l4)所以l4不在 l2 之后

我们都有 !(l2 < l4) && !(l4 < l2)所以有等效，我们不插入l4 .

3) Why does below code not work for removing overlapping intervals?

我不明白你在哪里使用它。