标签：class perfect follow gap lap unit example sum represent
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.
这个题我甚至不知道该怎么总结。
难就难在从这个题抽象出一种解法，看了别人的答案和思路= =然而没有归类总结到某种类型，这题相当于背了个题。。。
简单的说，除了最外面的4个点，所有的点都会2次2次的出现，如果有覆盖，覆盖进去的点就不是成对出现了。
最外面4个点围的面积等于所有小矩形面积的和。
就用这2个判断就行了。
判断成对的点用的SET，单次出现添加，第二次出现删除。。这样最后应该都删掉，SET里只剩下4个最外面的点。
剩下的就是判断最外点，不停地更新。。
Consider how the corners of all rectangles appear in the large rectangle if there‘s a perfect rectangular cover.
Rule1: The local shape of the corner has to follow one of the three following patterns
For each point being a corner of any rectangle, it should appear even times except the 4 corners of the large rectangle. So we can put those points into a hash map and remove them if they appear one more time.
At the end, we should only get 4 points.
Rule2: the large rectangle area should be equal to the sum of small rectangles
public class Solution {
public boolean isRectangleCover(int[][] rectangles) {
if (rectangles==null || rectangles.length==0 || rectangles[0].length==0) return false;
int subrecAreaSum = 0; //sum of subrectangle‘s area
int x1 = Integer.MAX_VALUE; //large rectangle bottom left x-axis
int y1 = Integer.MAX_VALUE; //large rectangle bottom left y-axis
int x2 = Integer.MIN_VALUE; //large rectangle top right x-axis
int y2 = Integer.MIN_VALUE; //large rectangle top right y-axis
HashSet<String> set = new HashSet<String>(); // store points
for(int[] rec : rectangles) {
//check if it has large rectangle‘s 4 points
x1 = Math.min(x1, rec[0]);
y1 = Math.min(y1, rec[1]);
x2 = Math.max(x2, rec[2]);
y2 = Math.max(y2, rec[3]);
//calculate sum of subrectangles
subrecAreaSum += (rec[2]-rec[0]) * (rec[3] - rec[1]);
//store this rectangle‘s 4 points into hashSet
String p1 = Integer.toString(rec[0]) + "" + Integer.toString(rec[1]);
String p2 = Integer.toString(rec[0]) + "" + Integer.toString(rec[3]);
String p3 = Integer.toString(rec[2]) + "" + Integer.toString(rec[1]);
String p4 = Integer.toString(rec[2]) + "" + Integer.toString(rec[3]);
if (!set.add(p1)) set.remove(p1);
if (!set.add(p2)) set.remove(p2);
if (!set.add(p3)) set.remove(p3);
if (!set.add(p4)) set.remove(p4);
}
if (set.size()!=4 || !set.contains(x1+""+y1) || !set.contains(x1+""+y2) || !set.contains(x2+""+y1) || !set.contains(x2+""+y2))
return false;
return subrecAreaSum == (x2-x1) * (y2-y1);
}
}
存为字符串 //store this rectangle‘s 4 points into hashSet
String p1 = Integer.toString(rec[0]) + "" + Integer.toString(rec[1]);
boolean |
add(E e) 如果此 set 中尚未包含指定元素，则添加指定元素。 |
标签：class perfect follow gap lap unit example sum represent
原文地址：http://www.cnblogs.com/apanda009/p/7150028.html
