218. The Skyline problem

A city's skyline is the outer contour of the silhouette formed by all the buildings in then city then viewed from a Distan Ce. Now suppose you is given the locations and height of all the buildings as shown in a cityscape photo (figure A), write a Program to output the skyline formed by these buildings collectively (Figure B).

The geometric information of each building was represented by a triplet of integers [Li, Ri, Hi] , where and is the Li Ri x Coordinates of the left and right edge of the ith building, respectively, and are its Hi height. It's guaranteed that 0 ≤ Li, Ri ≤ INT_MAX , 0 < Hi ≤ INT_MAX and Ri - Li > 0 . Assume all buildings is perfect rectangles grounded on a absolutely flat surface at height 0.

For instance, the dimensions of all buildings in Figure A is recorded as: [ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ]  .

The output is a list of "key points" (Red dots in Figure B) in the format of that [ [x1,y1], [x2, y2], [x3, y3], ... ] uniquely defines a skyline. A key point was the left endpoint of a horizontal line segment. Note the last key point, where the rightmost building ends, was merely used to mark the termination of the skyline, an D always have zero height. Also, the ground in between any and adjacent buildings should is considered part of the Skyline Contour.

For instance, the skyline in Figure B should is represented as: [ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ] .

Notes:

• The number of buildings in all input list is guaranteed to being in the range [0, 10000] .
• The input list is already sorted in ascending order by the left X position Li .
• The output list must is sorted by the X position.
• There must is no consecutive horizontal lines of equal height in the output skyline. For instance, was not [...[2 3], [4 5], [7 5], [11 5], [12 7]...] acceptable; The three lines of height 5 should being merged into one in the final output as such:[...[2 3], [4 5], [12 7], ...]

Points:skyline mean set of those Points who change hight

1 /*2 * Solution 1 Heap--Priorityqueue3  */4  Public classSolution {5      Publiclist<int[]> Getskyline (int[] buildings) {6list<int[]> list =Newlinkedlist<int[]>();7list<int[]> height =Newlinkedlist<int[]>();8         9          for(int[] b:buildings) {TenHeight.add ( new int[]{b[0],-b[2]}); OneHeight.add (New int[]{b[1], b[2]}); A         } -          -         Collections.sort(Height, (A, B), {  the             if(a[0]! = b[0]) -                 returnA[0]-b[0];//Small to large -             returnA[1]-b[1];// Low-to-high -  }); +          -Priorityqueue<integer> PQ =NewPriorityqueue<integer> ((A, B)(b- a)); +          int pre = 0;  A         pq.offer (0);  at          -          for(int[] h:height) { -             if(H[1] < 0) { -Pq.offer (-h[1]); -}Else { -Pq.Remove(h[1]); in             } -             intCur =Pq.peek (); to             if(Pre! =cur) { //  change of Height point +List.add (New int[]{h[0], cur}); -Pre =cur; the             } *         } $         returnlist;Panax Notoginseng     } -}

1 /*2 * Solution 2 Divide and conquer--merge3  */4  Public classSolution {5      Publiclist<int[]> Getskyline (int[] buildings) {6         if(Buildings.length = = 0)7             return Newlinkedlist<int[]>();8         returnRecurskyline (buildings, 0, buildings.length-1);9     }Ten      One     Private LinkedList<int[]> Recurskyline (int[] Buildings,intSintt) { A         if(s = =t) { -linkedlist<int[]> ret =Newlinkedlist<int[]>(); -Ret.add (New int[]{buildings[s][0], buildings[s][2]});//Left top theRet.add (New int[]{buildings[s][1], 0});//Right , bottom. -             returnret; -}Else { -             intMID = (T-s)/2 +s; +             returnMerge (Recurskyline (buildings, S, mid), Recurskyline (buildings, mid + 1, T)); -         } +     } A      at     Private LinkedList<int[]> Merge (linkedlist<int[]> L1, linkedlist<int[]>L2) { -         intH = 0; -         intx = 0; -         intH1 = 0, h2 = 0; -linkedlist<int[]> ret =Newlinkedlist<int[]>(); -          in          while(!l1.isempty () &&!L2.isempty ()) { -             if(L1.getfirst () [0] < L2.getfirst () [0]) { tox = L1.GetFirst() [0]; //GetFirst, Removefirst, getlast based on LinkedList, List  does not have these functions +H1 = L1.getfirst () [1]; -h =Math.max (H1, H2); the L1.removefirst (); *}Else if(L1.getfirst () [0] > L2.getfirst () [0]) { $x = L2.getfirst () [0];Panax NotoginsengH2 = L2.getfirst () [1]; -h =Math.max (H1, H2); the L2. Removefirst (); +}Else { Ax = L1.getfirst () [0]; theH1 = L1.getfirst () [1]; +H2 = L2.getfirst () [1]; -h =Math.max (H1, H2); $ L1.removefirst (); $ L2.removefirst (); -             } -              the             if(ret.size () = = 0 | | ret.getlast () [1]! = h) {//Height Change point -Ret.add (New int[]{x, H}];Wuyi             } the         } -          Wu Ret.addall (L1); - Ret.addall (L2); About         returnret; $     } -}

218. The Skyline problem