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[Swift]LeetCode766. 托普利茨矩阵 | Toeplitz Matrix

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A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.

Now given an M x N matrix, return True if and only if the matrix is Toeplitz

Example 1:

Input:
matrix = [
  [1,2,3,4],
  [5,1,2,3],
  [9,5,1,2]
]
Output: True
Explanation:
In the above grid, the diagonals are:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
In each diagonal all elements are the same, so the answer is True.

Example 2:

Input:
matrix = [
  [1,2],
  [2,2]
]
Output: False
Explanation:
The diagonal "[1, 2]" has different elements.

Note:

    1. matrix will be a 2D array of integers.
    2. matrix will have a number of rows and columns in range [1, 20].
    3. matrix[i][j] will be integers in range [0, 99].

如果一个矩阵的每一方向由左上到右下的对角线上具有相同元素,那么这个矩阵是托普利茨矩阵

给定一个 M x N 的矩阵,当且仅当它是托普利茨矩阵时返回 True

示例 1:

输入: 
matrix = [
  [1,2,3,4],
  [5,1,2,3],
  [9,5,1,2]
]
输出: True
解释:
在上述矩阵中, 其对角线为:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]"。
各条对角线上的所有元素均相同, 因此答案是True。

示例 2:

输入:
matrix = [
  [1,2],
  [2,2]
]
输出: False
解释: 
对角线"[1, 2]"上的元素不同。

说明:

  1.  matrix 是一个包含整数的二维数组。
  2. matrix 的行数和列数均在 [1, 20]范围内。
  3. matrix[i][j] 包含的整数在 [0, 99]范围内。

进阶:

  1. 如果矩阵存储在磁盘上,并且磁盘内存是有限的,因此一次最多只能将一行矩阵加载到内存中,该怎么办?
  2. 如果矩阵太大以至于只能一次将部分行加载到内存中,该怎么办?

Runtime: 28 ms
Memory Usage: 19 MB
 1 class Solution {
 2     func isToeplitzMatrix(_ matrix: [[Int]]) -> Bool {
 3         for i in 0..<matrix.count - 1
 4         {
 5             for j in 0..<matrix[i].count - 1
 6             {
 7                 if matrix[i][j] != matrix[i + 1][j + 1]
 8                 {
 9                     return false
10                 }
11             }
12         }
13         return true
14     }
15 }

28ms

 1 class Solution {
 2     func isToeplitzMatrix(_ matrix: [[Int]]) -> Bool {
 3         var temp: [Int] = matrix[0]
 4         for i in 1..<matrix.count {
 5             let smallMatrix = matrix[i]
 6             for j in 0..<smallMatrix.count {
 7                 if j == 0 {
 8                     temp.insert(smallMatrix[j], at: 0)
 9                 } else {
10                     if smallMatrix[j] != temp[j] {
11                         return false
12                     }
13                 }
14             }
15         }
16         return true
17     }
18 }

32ms

 1 class Solution {
 2     func isToeplitzMatrix(_ matrix: [[Int]]) -> Bool {
 3         var result = true
 4 
 5         for i in 0..<matrix.count {
 6             for j in 0..<matrix[0].count {
 7                 if i > 0 && j > 0 {
 8                     break;
 9                 }
10                 var pointR = i + 1
11                 var pointC = j + 1
12 
13                 let val1 = matrix[i][j];
14                 let count = matrix[0].count
15 
16                 while pointR < matrix.count {
17                     if pointC < count {
18                         let val2 = matrix[pointR][pointC]
19                         if val1 != val2 {
20                             result = false;
21                         }
22                         pointR += 1;
23                         pointC += 1;
24                     } else {
25                         break;
26                     }
27                 }
28             }
29         }
30         return result;
31     }
32 }

68ms

 1 class Solution {
 2     func isToeplitzMatrix(_ matrix: [[Int]]) -> Bool {
 3 for i in 0..<matrix[0].count{
 4             var val = -1
 5             var rowIndex = 0
 6             var columnIndex = i
 7             
 8             while rowIndex < matrix.count && columnIndex < matrix[0].count{
 9                 if val != -1 && val != matrix[rowIndex][columnIndex]{
10                     return false
11                 }
12                 val = matrix[rowIndex][columnIndex]
13                 rowIndex += 1
14                 columnIndex += 1
15             }
16         }
17        
18         for i in 0..<matrix.count{
19             var val = -1
20             
21             var rowIndex = i
22             var columnIndex = 0
23             
24             while rowIndex < matrix.count && columnIndex < matrix[0].count{
25                 if val != -1 && val != matrix[rowIndex][columnIndex]{
26                     return false
27                 }
28                 val = matrix[rowIndex][columnIndex]
29                 rowIndex += 1
30                 columnIndex += 1
31             }            
32         }
33         return true
34     }
35 }

88ms

 1 class Solution {
 2     func isToeplitzMatrix(_ matrix: [[Int]]) -> Bool {
 3         guard matrix.count > 0 else {
 4             return false
 5         }
 6         if matrix.count == 1 {
 7             return true
 8         }
 9         
10         var row = matrix[0]
11         for i in 1..<matrix.count {
12             row = matrix[i-1]
13             if !matrix[i][1...].elementsEqual(row[0..<row.count - 1]) {
14                 return false
15             }
16         }
17         return true
18     }
19 }

 

posted @ 2019-03-14 20:49  为敢技术  阅读(259)  评论(0编辑  收藏  举报