重叠矩形的合并

问题描述：输入一组矩形，这些矩形的边与坐标轴平行，要求合并重叠的矩形并输出合并后的结果。

面试的时候虽然感觉可以先排序再搜索，但没想到具体的做法，所以就直接暴搜了。

import java.util.*;
class Main {
    public static void main(String[] args){
        List input = new LinkedList();
        boolean merged = true;
        while(merged){
            merged = false;
            for(int i = 0; i < input.size(); i++){
                for(int j = i + 1; j < input.size(); i++){
                    AABox box1 = input.get(i), box2 = input.get(j);
                    if(box1.IsOverlap(box2)){
                        input.set(i, box1.Merge(box2));
                        input.remove(j);
                        j--;
                        merged = true;
                    }
                }
            }
        }
        System.out.println(intput);
    }
}

这一做法在最坏情况下，时间复杂度是 O(n^3)。最坏情况如下图所示。

在上述极端情况下的时间复杂度为 O(n)。
（本来排序之后想用BFS的，后来发现实际情况比较简单，没必要BFS）

定义结构体Position和对象Box，Box代表矩形，Position用来包装 x 或者 y 值。分别维护2个Position链表，其中1个链表的元素是包装 x 值的Position，另1个则是 y 值。x 值的有序链表如下图所示。

struct Position{
public:
    int val;
    Position* prev, *next;
    Box* box;
    Position(int val){
        this->val = val;
        this->prev = NULL;
        this->next = NULL;
        this->box = NULL;
    }

    void setBox(Box * box){
        this->box = box;
    }
};

class Box{
public:
    Position *x0, *x1, *y0, *y1;
    bool deleted;
    Box(Position* in_x0, Position* in_y0, Position* in_x1, Position* in_y1) :
        x0(in_x0), y0(in_y0), x1(in_x1), y1(in_y1)
    {
        deleted = false;
        x0->setBox(this);
        y0->setBox(this);
        x1->setBox(this);
        y1->setBox(this);
    }
//    bool isOverlap(Box other);
//    Box merge(Box other);
};

举例：在上图的链表中，遍历 box1.x0 和 box1.x1 之间的节点，找到 节点2，并根据节点2的box指针找到所属的box2；比较box1和box2的y值，以判断box1和box2是否重合。

Position *ps = box1->x0->next;
while (ps->box != box1) {
     Box *box2 = ps->box;
     // Is overlap?
     if (box1->y0->val < box2->y0->val && box2->y0->val < box1->y1->val ||
         box1->y0->val < box2->y1->val && box2->y1->val < box1->y1->val ||
         box2->y0->val < box1->y0->val && box1->y1->val < box2->y1->val) {
         //merge

更新box1，将box2标记为删除状态，同时更新2个Position有序链表。
举例：上图中 box1.x1 = 3, box2.x1 = 5，所以将box1.x1的指针修改为指向节点5，并删除节点3

//merge
 if (box1->x0->val > box2->x0->val) {
     deleteNode(box1->x0);
     box1->x0 = box2->x0;
     box2->x0->box = box1;
 } else
     deleteNode(box2->x0);

 if (box1->x1->val < box2->x1->val) {
     deleteNode(box1->x1);
     box1->x1 = box2->x1;
     box2->x1->box = box1;
 } else
     deleteNode(box2->x1);

 if (box1->y0->val > box2->y0->val) {
     deleteNode(box1->y0);
     box1->y0 = box2->y0;
     box2->y0->box = box1;
 } else
     deleteNode(box2->y0);

 if (box1->y1->val < box2->y1->val) {
     deleteNode(box1->y1);
     box1->y1 = box2->y1;
     box2->y1->box = box1;
 } else
     deleteNode(box2->y1);
 //merge

#include 
#include 
#include 
#include 

using namespace std;

class Box;

struct Position{
public:
    int val;
    Position* prev, *next;
    Box* box;
    Position(int val){
        this->val = val;
        this->prev = NULL;
        this->next = NULL;
        this->box = NULL;
    }

    void setBox(Box * box){
        this->box = box;
    }
};

class Box{
public:
    Position *x0, *x1, *y0, *y1;
    bool deleted;
    Box(Position* in_x0, Position* in_y0, Position* in_x1, Position* in_y1) :
        x0(in_x0), y0(in_y0), x1(in_x1), y1(in_y1)
    {
        deleted = false;
        x0->setBox(this);
        y0->setBox(this);
        x1->setBox(this);
        y1->setBox(this);
    }
//    bool isOverlap(Box other);
//    Box merge(Box other);
};

bool cmp(int* p1, int* p2){
    return *p1 < *p2;
}

void insertSort(Position & h, Position & t, Position & n){
    Position * i = &t;
    while (&h != i && ((*i).val > n.val || &t == i)){
        i = (*i).prev;
    }
    n.next = (*i).next;
    n.prev = i;
    (*i).next->prev = &n;
    (*i).next = &n;
}

void deleteNode(Position * n) {
    n->prev->next = n->next;
    n->next->prev = n->prev;
    delete n;
}

int main() {
    list in;
    Position xHead(0), xTail(0), yHead(0), yTail(0);
    xHead.next = &xTail;
    xTail.prev = &xHead;
    yHead.next = &yTail;
    yTail.prev = &yHead;

    int n;
    cin >> n;
    while (n--) {
        int x0, y0, x1, y1;
        cin >> x0 >> y0 >> x1 >> y1;

        Position* px0 = new Position(x0);
        Position* py0 = new Position(y0);
        Position* px1 = new Position(x1);
        Position* py1 = new Position(y1);
        insertSort(xHead, xTail,*px0);
        insertSort(xHead, xTail,*px1);
        insertSort(yHead, yTail,*py0);
        insertSort(yHead, yTail,*py1);

        Box* box = new Box(px0, py0, px1, py1);
        in.push_back(box);
    }
    


    list::iterator it;
    for(it = in.begin(); it != in.end(); it++){
        if((*it)->deleted)
            continue;
            

            
        Box * box1 = *it;
        while(!box1->deleted &&
              box1->x0->next->box != box1 && box1->y0->next->box != box1) {

            Position *ps = box1->x0->next;
            while (ps->box != box1) {
                Box *box2 = ps->box;
                // Is overlap?
                if (box1->y0->val < box2->y0->val && box2->y0->val < box1->y1->val ||
                    box1->y0->val < box2->y1->val && box2->y1->val < box1->y1->val ||
                    box2->y0->val < box1->y0->val && box1->y1->val < box2->y1->val) {
                    //merge
                    if (box1->x0->val > box2->x0->val) {
                        deleteNode(box1->x0);
                        box1->x0 = box2->x0;
                        box2->x0->box = box1;
                    } else
                        deleteNode(box2->x0);

                    if (box1->x1->val < box2->x1->val) {
                        deleteNode(box1->x1);
                        box1->x1 = box2->x1;
                        box2->x1->box = box1;
                    } else
                        deleteNode(box2->x1);

                    if (box1->y0->val > box2->y0->val) {
                        deleteNode(box1->y0);
                        box1->y0 = box2->y0;
                        box2->y0->box = box1;
                    } else
                        deleteNode(box2->y0);

                    if (box1->y1->val < box2->y1->val) {
                        deleteNode(box1->y1);
                        box1->y1 = box2->y1;
                        box2->y1->box = box1;
                    } else
                        deleteNode(box2->y1);
                    //merge

                    //delete box2
                    box2->deleted = true;
                }

                ps = ps->next;
            }
        }
    }

    for(it = in.begin(); it != in.end(); it++) {
        if((*it)->deleted)
            it = in.erase(it);
    }

    for(it = in.begin(); it != in.end(); it++){
        cout << (*it)->x0->val << " " << (*it)->y0->val << " " << (*it)->x1->val << " " << (*it)->y1->val << endl;
    }
}

有人用并查集解决此问题（链接），但个人感觉没有必要使用并查集。