391. Perfect Rectangle

本文介绍了一种算法,用于判断一组轴对齐的矩形是否完全覆盖一个矩形区域,且无重叠和空隙。通过遍历矩形集合,确定最大轮廓,并使用哈希集记录顶点出现次数,最终验证覆盖的有效性。

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.

在这里插入图片描述

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.

在这里插入图片描述

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.

在这里插入图片描述

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.

在这里插入图片描述

思路:遍历一遍所有的矩形,得到所有矩形的最大轮廓,在遍历过程中记录矩形的四个顶点构成的坐标出现的次数,如果是完美矩形,则这些顶点出现的次数出轮廓顶点外会出现偶数次,轮廓顶点出现一次
public boolean isRectangleCover(int[][] rectangles) {
       int top=rectangles[0][2];
       int right=rectangles[0][3];
       int left=rectangles[0][1];
       int bottom=rectangles[0][0];
       HashSet<String> set = new HashSet<String>();
       int area=0;
       for(int[] rectangle:rectangles){
           top=Math.max(top,rectangle[2]);
           right=Math.max(right,rectangle[3]);
           left=Math.min(left,rectangle[1]);
           bottom=Math.min(bottom,rectangle[0]);
           String s1 = rectangle[0] + " " + rectangle[1];
           String s2 = rectangle[0] + " " + rectangle[3];
           String s3 = rectangle[2] + " " + rectangle[3];
           String s4 = rectangle[2] + " " + rectangle[1];
           area+=(rectangle[2]-rectangle[0])*(rectangle[3]-rectangle[1]);
           if (!set.add(s1)) set.remove(s1);
           if (!set.add(s2)) set.remove(s2);
           if (!set.add(s3)) set.remove(s3);
           if (!set.add(s4)) set.remove(s4);

       }
       if (!set.contains(bottom + " " + left) || !set.contains(bottom + " " + right) || !set.contains(top + " " + left) || !set.contains(top + " " + right) || set.size() != 4) return false;
       return area==(top-bottom)*(right-left);
   }
Implement the classic method for composing secret messages called a square code. Given an English text, output the encoded version of that text. First, the input is normalized: the spaces and punctuation are removed from the English text and the message is down-cased. Then, the normalized characters are broken into rows. These rows can be regarded as forming a rectangle when printed with intervening newlines. For example, the sentence ```text "If man was meant to stay on the ground, god would have given us roots." ``` is normalized to: ```text "ifmanwasmeanttostayonthegroundgodwouldhavegivenusroots" ``` The plaintext should be organized into a rectangle as square as possible. The size of the rectangle should be decided by the length of the message. If `c` is the number of columns and `r` is the number of rows, then for the rectangle `r` x `c` find the smallest possible integer `c` such that: - `r * c >= length of message`, - and `c >= r`, - and `c - r <= 1`. Our normalized text is 54 characters long, dictating a rectangle with `c = 8` and `r = 7`: ```text "ifmanwas" "meanttos" "tayonthe" "groundgo" "dwouldha" "vegivenu" "sroots " ``` The coded message is obtained by reading down the columns going left to right. The message above is coded as: ```text "imtgdvsfearwermayoogoanouuiontnnlvtwttddesaohghnsseoau" ``` Output the encoded text in chunks that fill perfect rectangles `(r X c)`, with `c` chunks of `r` length, separated by spaces. For phrases that are `n` characters short of the perfect rectangle, pad each of the last `n` chunks with a single trailing space. ```text "imtgdvs fearwer mayoogo anouuio ntnnlvt wttddes aohghn sseoau " ``` 上述为题目要求,请用C语言实现,函数声明如下: char *ciphertext(const char *input)
最新发布
08-13
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