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An image is represented by a binary matrix with 0
as a white pixel and 1
as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y)
of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.git
For example, given the following image:github
[ "0010", "0110", "0100" ]
and x = 0
, y = 2
,数组
Return 6
.微信
图像由一个二进制矩阵表示,其中0为白色像素,1为黑色像素。黑色像素链接,即只有一个黑色区域。像素水平和垂直链接。给定一个黑色像素的位置(x, y),返回包围全部黑色像素的最小(轴对齐)矩形的区域。函数
例如,给出如下图像:测试
[ "0010", "0110", "0100" ]
X=0,Y=2,spa
返回6。code
Solution:htm
1 class Solution { 2 func minArea(_ image:inout [String],_ x:Int,_ y:Int) -> Int { 3 var m:Int = image.count 4 var n:Int = image[0].count 5 var up:Int = binary_search(&image, true, 0, x, 0, n, true) 6 var down:Int = binary_search(&image, true, x + 1, m, 0, n, false) 7 var left:Int = binary_search(&image, false, 0, y, up, down, true) 8 var right:Int = binary_search(&image, false, y + 1, n, up, down, false) 9 return (right - left) * (down - up) 10 } 11 12 // Binary Search 13 func binary_search(_ image:inout [String],_ h:Bool,_ i:Int,_ j:Int,_ low:Int,_ high:Int,_ opt:Bool) -> Int 14 { 15 var i = i 16 var j = j 17 while (i < j) 18 { 19 var k:Int = low 20 var mid:Int = (i + j) / 2 21 while (k < high && (h ? image[mid][k] : image[k][mid]) == "0") 22 { 23 k += 1 24 } 25 if (k < high) == opt{ j = mid } 26 else{ i = mid + 1 } 27 } 28 return i 29 } 30 } 31 32 //String扩展 33 extension String { 34 //subscript函数能够检索数组中的值 35 //直接按照索引方式截取指定索引的字符 36 subscript (_ i: Int) -> Character { 37 //读取字符 38 get {return self[index(startIndex, offsetBy: i)]} 39 } 40 }
点击:Playground测试
1 var sol = Solution() 2 var arr:[String] = ["0010","0110","0100"] 3 var x:Int = 0 4 var y:Int = 2 5 print(sol.minArea(&arr,x,y)) 6 //Print 6