[Swift]LeetCode240.搜索二维矩阵II|Searcha2DMatrixII

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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：

每行的元素从左到右升序排列。
每列的元素从上到下升序排列。

示例:

现有矩阵 matrix 如下：

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

给定 target = 5，返回 true。

给定 target = 20，返回 false。

328ms

 1 class Solution {
 2     func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
 3         // start at top right
 4         if matrix.count < 1 || matrix[0].count < 1 { return false }
 5         
 6         let rows = matrix.count, cols = matrix[0].count
 7         var r = 0, c = cols - 1
 8         
 9         while c >= 0, r < rows {
10             let num = matrix[r][c]
11             if num == target {
12                 return true
13             } else if num < target {
14                 r += 1
15             } else {
16                 c -= 1
17             }
18         }
19         
20         return false
21     }
22 }

332ms

 1 class Solution {
 2     func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
 3     guard matrix.count > 0, matrix[0].count > 0 else {
 4         return false
 5     }
 6     
 7     let m = matrix.count
 8     let n = matrix[0].count
 9     
10     if target < matrix[0][0] || target > matrix[m-1][n-1] {
11         return false
12     }
13     
14     var i = m-1, j = 0
15     
16     while i >= 0, j < n {
17         if matrix[i][j] == target {
18             return true
19         } else if matrix[i][j] > target {
20             i -= 1
21         } else {
22             j += 1
23         }
24     }
25     
26     return false
27     }
28 }

332ms

 1 class Solution {
 2     func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
 3         guard !matrix.isEmpty && !matrix[0].isEmpty else {
 4             return false
 5         }
 6 
 7         var row = 0
 8         var column = matrix[0].count - 1
 9         while column >= 0 && row < matrix.count {
10             if matrix[row][column] > target {
11                 column -= 1
12             } else if matrix[row][column] < target {
13                 row += 1
14             } else {
15                 return true
16             }
17         }
18 
19         return false
20     }
21 }

340ms

 1 class Solution {
 2     func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
 3 
 4         guard matrix.count > 0, matrix[0].count > 0 else { return false }
 5         var m = matrix.count, n = matrix[0].count
 6         var i = m-1, j = 0
 7         while true {
 8             if matrix[i][j] == target { return true }
 9             else if matrix[i][j] < target { j += 1 }
10             else {
11                 i -= 1
12             }
13             if i < 0 || j >= n { return false }
14         }
15         return false
16     }
17 }

348ms

 1 class Solution {
 2     func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
 3         for i in 0..<matrix.count {
 4             for j in stride(from: matrix[i].count, to: 0, by: -1) {
 5                     if matrix[i][j - 1] == target {
 6                         return true
 7                     }
 8             }
 9         }
10         
11         return false 
12     }
13 }