Here I'm posting my Java code for the skyline problem. If you have time and would like to review, please do so, I'd appreciate that.
Problem
A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are given the locations and height of all the buildings as shown on a cityscape photo (Figure A), write a program to output the skyline formed by these buildings collectively (Figure B).
Buildings Skyline Contour The geometric information of each building is represented by a triplet of integers [Li, Ri, Hi], where Li and Ri are the x coordinates of the left and right edge of the ith building, respectively, and Hi is its height. It is guaranteed that 0 ≤ Li, Ri ≤ INT_MAX, 0 < Hi ≤ INT_MAX, and Ri - Li > 0. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.
For instance, the dimensions of all buildings in Figure A are 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 [ [x1,y1], [x2, y2], [x3, y3], ... ] that uniquely defines a skyline. A key point is the left endpoint of a horizontal line segment. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.
For instance, the skyline in Figure B should be represented as:[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ].
Notes:
The number of buildings in any input list is guaranteed to be in the range [0, 10000]. The input list is already sorted in ascending order by the left x position Li. The output list must be sorted by the x position. There must be no consecutive horizontal lines of equal height in the output skyline. For instance, [...[2 3], [4 5], [7 5], [11 5], [12 7]...] is not acceptable; the three lines of height 5 should be merged into one in the final output as such: [...[2 3], [4 5], [12 7], ...]
Code
import java.util.Arrays;
import java.util.List;
import java.util.LinkedList;
import java.util.TreeMap;
import java.util.Collections;
class Solution {
public List<List<Integer>> getSkyline(int[][] buildings) {
List<List<Integer>> heights = new LinkedList<List<Integer>>();
TreeMap<Integer, Integer> heightTreeMap = new TreeMap<Integer, Integer>();
List<List<Integer>> contour = new LinkedList<List<Integer>>();
int lastHeight = 0;
for (int[] building : buildings) {
heights.add(Arrays.asList(building[0], -building[2]));
heights.add(Arrays.asList(building[1], building[2]));
}
Collections.sort(heights, (a, b) -> (
a.get(0).intValue() != b.get(0).intValue() ? a.get(0) - b.get(0) : a.get(1) - b.get(1)
));
heightTreeMap.put(0, 1);
for (List<Integer> val : heights) {
int height = val.get(1);
if (height < 0)
heightTreeMap.put(-height, -~heightTreeMap.getOrDefault(-height, 0));
else if (heightTreeMap.getOrDefault(height, 0) > 1)
heightTreeMap.put(height, heightTreeMap.get(height) - 1);
else
heightTreeMap.remove(height);
if (heightTreeMap.lastKey() != lastHeight) {
lastHeight = heightTreeMap.lastKey();
contour.add(Arrays.asList(val.get(0), lastHeight));
}
}
return contour;
}
}
On LeetCode, we are only allowed to change the argument names and brute force algorithms are discouraged, usually fail with TLE (Time Limit Error) or MLE (Memory Limit Error).
LeetCode's Solution - Not for review
class Solution {
/**
* Divide-and-conquer algorithm to solve skyline problem,
* which is similar with the merge sort algorithm.
*/
public List<List<Integer>> getSkyline(int[][] buildings) {
int n = buildings.length;
List<List<Integer>> output = new ArrayList<List<Integer>>();
// The base cases
if (n == 0) return output;
if (n == 1) {
int xStart = buildings[0][0];
int xEnd = buildings[0][1];
int y = buildings[0][2];
output.add(new ArrayList<Integer>() {{add(xStart); add(y); }});
output.add(new ArrayList<Integer>() {{add(xEnd); add(0); }});
// output.add(new int[]{xStart, y});
// output.add(new int[]{xEnd, 0});
return output;
}
// If there is more than one building,
// recursively divide the input into two subproblems.
List<List<Integer>> leftSkyline, rightSkyline;
leftSkyline = getSkyline(Arrays.copyOfRange(buildings, 0, n / 2));
rightSkyline = getSkyline(Arrays.copyOfRange(buildings, n / 2, n));
// Merge the results of subproblem together.
return mergeSkylines(leftSkyline, rightSkyline);
}
/**
* Merge two skylines together.
*/
public List<List<Integer>> mergeSkylines(List<List<Integer>> left, List<List<Integer>> right) {
int nL = left.size(), nR = right.size();
int pL = 0, pR = 0;
int currY = 0, leftY = 0, rightY = 0;
int x, maxY;
List<List<Integer>> output = new ArrayList<List<Integer>>();
// while we're in the region where both skylines are present
while ((pL < nL) && (pR < nR)) {
List<Integer> pointL = left.get(pL);
List<Integer> pointR = right.get(pR);
// pick up the smallest x
if (pointL.get(0) < pointR.get(0)) {
x = pointL.get(0);
leftY = pointL.get(1);
pL++;
}
else {
x = pointR.get(0);
rightY = pointR.get(1);
pR++;
}
// max height (i.e. y) between both skylines
maxY = Math.max(leftY, rightY);
// update output if there is a skyline change
if (currY != maxY) {
updateOutput(output, x, maxY);
currY = maxY;
}
}
// there is only left skyline
appendSkyline(output, left, pL, nL, currY);
// there is only right skyline
appendSkyline(output, right, pR, nR, currY);
return output;
}
/**
* Update the final output with the new element.
*/
public void updateOutput(List<List<Integer>> output, int x, int y) {
// if skyline change is not vertical -
// add the new point
if (output.isEmpty() || output.get(output.size() - 1).get(0) != x)
output.add(new ArrayList<Integer>() {{add(x); add(y); }});
// if skyline change is vertical -
// update the last point
else {
output.get(output.size() - 1).set(1, y);
}
}
/**
* Append the rest of the skyline elements with indice (p, n)
* to the final output.
*/
public void appendSkyline(List<List<Integer>> output, List<List<Integer>> skyline,
int p, int n, int currY) {
while (p < n) {
List<Integer> point = skyline.get(p);
int x = point.get(0);
int y = point.get(1);
p++;
// update output
// if there is a skyline change
if (currY != y) {
updateOutput(output, x, y);
currY = y;
}
}
}
}
