LeetCode - Perfect Rectangle
作者:互联网
Given N axis-aligned rectangles where N > 0, determine if they all together form an exact cover of a rectangular region. Each rectangle is represented as a bottom-left point and a top-right point. For example, a unit square is represented as [1,1,2,2]. (coordinate of bottom-left point is (1, 1) and top-right point is (2, 2)). Example 1: rectangles = [ [1,1,3,3], [3,1,4,2], [3,2,4,4], [1,3,2,4], [2,3,3,4] ] Return true. All 5 rectangles together form an exact cover of a rectangular region. Example 2: rectangles = [ [1,1,2,3], [1,3,2,4], [3,1,4,2], [3,2,4,4] ] Return false. Because there is a gap between the two rectangular regions. Example 3: rectangles = [ [1,1,3,3], [3,1,4,2], [1,3,2,4], [3,2,4,4] ] Return false. Because there is a gap in the top center. Example 4: rectangles = [ [1,1,3,3], [3,1,4,2], [1,3,2,4], [2,2,4,4] ] Return false. Because two of the rectangles overlap with each other.
class Solution { // 1. Exact 4 corners // 2. Other shared points must occur evenly (cannot be odd) // 3. sum(area) = area of 4 corners enclosed areas private String delimiter = ","; public boolean isRectangleCover(int[][] rectangles) { Map<String, Integer> set = new HashMap<>(); Map<String, int[]> map2 = new HashMap<>(); int sum = 0; for (int i = 0; i < rectangles.length; i++) { sum = sum + (rectangles[i][2] - rectangles[i][0]) * (rectangles[i][3] - rectangles[i][1]); // bottom-left String blKey = generateKey(rectangles[i][0], rectangles[i][1]); if (set.containsKey(blKey)) { map2.remove(blKey); } else { map2.put(blKey, new int[]{rectangles[i][0], rectangles[i][1]}); } set.put(blKey, set.getOrDefault(blKey, 0) + 1); // top-right String trKey = generateKey(rectangles[i][2], rectangles[i][3]); if (set.containsKey(trKey)) { map2.remove(trKey); } else { map2.put(trKey, new int[]{rectangles[i][2], rectangles[i][3]}); } set.put(trKey, set.getOrDefault(trKey, 0) + 1); //top-left String tlKey = generateKey(rectangles[i][0], rectangles[i][3]); if (set.containsKey(tlKey)) { map2.remove(tlKey); } else { map2.put(tlKey, new int[]{rectangles[i][0], rectangles[i][3]}); } set.put(tlKey, set.getOrDefault(tlKey, 0) + 1); //bottom-right String brKey = generateKey(rectangles[i][2], rectangles[i][1]); if (set.containsKey(brKey)) { map2.remove(brKey); } else { map2.put(brKey, new int[]{rectangles[i][2], rectangles[i][1]}); } set.put(brKey, set.getOrDefault(brKey, 0) + 1); } if (map2.size() == 4) { for (String key: set.keySet()) { if (set.get(key) % 2 != 0 && set.get(key) != 1) { return false; } } int maxL = Integer.MIN_VALUE; int minL = Integer.MAX_VALUE; int maxW = Integer.MIN_VALUE; int minW = Integer.MAX_VALUE; for (String key : map2.keySet()) { if(map2.get(key)[0] > maxL) { maxL = map2.get(key)[0]; } if(map2.get(key)[0] < minL) { minL = map2.get(key)[0]; } if(map2.get(key)[1] > maxW) { maxW = map2.get(key)[1]; } if(map2.get(key)[1] < minW) { minW = map2.get(key)[1]; } } System.out.println("Sum: " + sum); System.out.println("Area: " + (maxL - minL) * (maxW - minW)); if(sum == (maxL - minL) * (maxW - minW)) { return true; } } return false; } private String generateKey(int i, int j) { return Integer.toString(i) + delimiter + Integer.toString(j); } }
标签:Perfect,set,get,int,map2,key,rectangles,LeetCode,Rectangle 来源: https://www.cnblogs.com/incrediblechangshuo/p/14436750.html