• 设为首页
  • 点击收藏
  • 手机版
    手机扫一扫访问
    迪恩网络手机版
  • 关注官方公众号
    微信扫一扫关注
    公众号

[Swift]LeetCode566.重塑矩阵|ReshapetheMatrix

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/9843320.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.

You're given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.

The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.

If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Example 1:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
Output: 
[[1,2,3,4]]
Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

 Example 2:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
Output: 
[[1,2],
 [3,4]]
Explanation:
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

 Note:

  1. The height and width of the given matrix is in range [1, 100].
  2. The given r and c are all positive.

在MATLAB中,有一个非常有用的函数 reshape,它可以将一个矩阵重塑为另一个大小不同的新矩阵,但保留其原始数据。

给出一个由二维数组表示的矩阵,以及两个正整数rc,分别表示想要的重构的矩阵的行数和列数。

重构后的矩阵需要将原始矩阵的所有元素以相同的行遍历顺序填充。

如果具有给定参数的reshape操作是可行且合理的,则输出新的重塑矩阵;否则,输出原始矩阵。

示例 1:

输入: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
输出: 
[[1,2,3,4]]
解释:
行遍历nums的结果是 [1,2,3,4]。新的矩阵是 1 * 4 矩阵, 用之前的元素值一行一行填充新矩阵。

示例 2:

输入: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
输出: 
[[1,2],
 [3,4]]
解释:
没有办法将 2 * 2 矩阵转化为 2 * 4 矩阵。 所以输出原矩阵。

注意:

  1. 给定矩阵的宽和高范围在 [1, 100]。
  2. 给定的 r 和 c 都是正数。

172ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         //创建一个二维数组      
 4         var res:[[Int]] = [[Int]](repeating: [Int](repeating: 0, count: c), count: r)
 5         let len:Int = nums.count
 6         if len == 0 || r * c != len * nums[0].count
 7         {
 8             return nums
 9         }
10         var count:Int = 0
11         for i in 0..<len
12         {
13             for j in 0..<nums[0].count
14             {
15                 res[count / c][count % c] = nums[i][j]
16                 count++
17             }    
18         }
19         return res
20     }
21 }
22 
23 /*扩展Int类,实现自增++、自减--运算符*/
24 extension Int{
25     //++前缀:先自增再执行表达示
26     static prefix func ++(num:inout Int) -> Int {
27         //输入输出参数num
28         num += 1
29         //返回加1后的数值
30         return num
31     }
32     //后缀++:先执行表达式后再自增
33     static postfix func ++(num:inout Int) -> Int {
34         //输入输出参数num
35         let temp = num
36         //num加1
37         num += 1
38         //返回加1前的数值
39         return temp
40     }
41     //--前缀:先自减再执行表达示
42     static prefix func --(num:inout Int) -> Int {
43         //输入输出参数num
44         num -= 1
45         //返回减1后的数值
46         return num
47     }
48     //后缀--:先执行表达式后再自减
49     static postfix func --(num:inout Int) -> Int {
50         //输入输出参数num
51         let temp = num
52         //num减1
53         num -= 1
54          //返回减1前的数值
55         return temp
56     }
57 }

36ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         if c * r != nums.count * nums[0].count{
 4             return nums
 5         }
 6         var res = Array(repeating: Array(repeating: 0,count: c) ,count: r)
 7         var count = 0
 8         for i in 0..<nums.count{
 9             for j in 0..<nums[0].count{
10                 res[count / c][count % c] = nums[i][j]
11                 count += 1
12             }
13         }
14         
15         return res
16     }
17 }

36ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         guard !nums.isEmpty else {
 4             return [[]]
 5         }
 6     
 7         guard nums.count * nums[0].count == r*c else {
 8             return nums
 9         }
10     
11         var result = [[Int]](repeating: [Int](repeating: 0, count: c), count: r)
12         var index = 0
13         let arr = nums.flatMap {$0}
14         for row in 0..<r {
15             for col in 0..<c {
16                 result[row][col] = arr[index]
17                 index += 1
18             }
19         }
20     
21         return result
22     }
23 }

40ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         let row = nums.count
 4         let column = nums[0].count
 5         
 6         if (row == r && column == c) || row * column != r * c {
 7             return nums
 8         }
 9         
10         var result = [[Int]]()
11         var index = 0
12         for _ in 0..<r {
13             var newRow = [Int]()
14             for _ in 0..<c {
15                 newRow.append(nums[index / column][index % column])
16                 index += 1
17             }
18             result.append(newRow)
19         }
20         return result
21     }
22 }

40ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         let rows = nums.count
 4         let cols = nums[0].count
 5         if r*c != rows*cols {
 6             return nums
 7         }
 8         var numbers = [Int]()
 9         for row in nums {
10             for item in row {
11                 numbers.append(item)
12             }
13         }
14         var reshaped = [[Int]]()
15         var rIndex = 0
16         var cIndex = 0
17         var row = [Int]()
18         numbers.forEach{ item in 
19             if cIndex == c {
20                 cIndex = 0
21                 rIndex += 1
22                 reshaped.append(row)
23                 row.removeAll()
24             }
25             row.append(item)
26             cIndex += 1
27         }
28         reshaped.append(row)
29         return reshaped
30     }
31 }

44ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         
 4         guard nums.count * nums[0].count == r * c else {
 5             return nums
 6         }
 7         
 8         var res = [[Int]]()
 9         var tmp = [Int]()   
10         var tmpCount = 0
11         
12         for num in nums {        
13             for n in num {             
14                 tmp.append(n)         
15                 if tmp.count >= c {
16                     res.append(tmp)
17                     tmp.removeAll()
18                 }
19             }
20         }
21         
22         return res
23     }
24 }

44ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         let numbers = Array(nums.joined())
 4         if numbers.count != r*c {
 5             return nums
 6         }
 7 
 8         var reshaped = Array(repeating:[Int](), count:r)
 9         for count in 0..<numbers.count {
10             reshaped[count/c].append(numbers[count])
11         }
12         return reshaped
13     }
14 }

48ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         let count = nums.count * nums[0].count;
 4         if count != r * c {
 5             return nums;
 6         }
 7         var result = Array<[Int]>()
 8         var temp = Array<Int>()
 9         
10         for array in nums {
11             for num in array {
12                 temp.append(num)
13                 if temp.count >= c {
14                     result.append(temp);
15                     temp.removeAll()
16                 }
17             }
18         }
19         return result;
20     }
21 }

48ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         var reshapedMatrix = [[Int]]()
 4         
 5         for _ in stride(from: 0, to: r, by: 1) {
 6             var tempRow = [Int]()
 7             for _ in stride(from: 0, to: c, by: 1) {
 8                 tempRow.append(0)
 9             }
10             reshapedMatrix.append(tempRow)
11         }
12         
13         let matrixSize = nums.count * nums[0].count
14         let reshapedMatrixSize = r * c
15         if matrixSize != reshapedMatrixSize {
16             return nums
17         }
18         for i in stride(from: 0, to: reshapedMatrixSize, by: 1) {
19             reshapedMatrix[i/c][i%c] = nums[i/nums[0].count][i%nums[0].count]
20         }
21         return reshapedMatrix
22     }
23 }

52ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3          if nums.count * nums[0].count != r * c {
 4             return nums
 5         }
 6         var temp = [Int](),result2 = [[Int]]()
 7         for itemArray in nums {
 8             temp += itemArray
 9         }
10         for i in 0..<r {
11             result2.append([Int](temp[i*c..<i*c+c]))
12         }
13         return result2
14     }
15 }

76ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         guard nums.count * nums.first!.count == r * c else {
 4             return nums
 5         }
 6         
 7         var newA = [[Int]]()
 8         var tmpA = [Int]()
 9         for subA in nums {
10             tmpA += subA
11         }
12         
13         for index in stride(from: 0, to: r * c, by: c) {
14             newA.append(Array(tmpA[index..<index+c]))
15         }
16         
17         return newA
18     }
19 }

84ms

 1 class Solution {
 2     func matrixReshape(_ nums: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {
 3         var new: [Int] = []
 4         for temp: [Int] in nums {
 5             for i: Int in temp {
 6                 new.append(i)
 7             }
 8         }
 9         if r * c > new.count {
10             return nums
11         }
12         var sq: [[Int]] = []
13         for j in 1...r {
14             var tempSq: [Int] = []
15             for a in 0..<c {
16                 tempSq.append(new[(j-1)*c+a])
17             }
18             sq.append(tempSq)
19         }
20         return sq
21     }
22 }

 


鲜花

握手

雷人

路过

鸡蛋
该文章已有0人参与评论

请发表评论

全部评论

专题导读
上一篇:
[Swift]LeetCode826.安排工作以达到最大收益|MostProfitAssigningWork发布时间:2022-07-13
下一篇:
swift如何控制view的显示与隐藏发布时间:2022-07-13
热门推荐
热门话题
阅读排行榜

扫描微信二维码

查看手机版网站

随时了解更新最新资讯

139-2527-9053

在线客服(服务时间 9:00~18:00)

在线QQ客服
地址:深圳市南山区西丽大学城创智工业园
电邮:jeky_zhao#qq.com
移动电话:139-2527-9053

Powered by 互联科技 X3.4© 2001-2213 极客世界.|Sitemap