[Swift]LeetCode240. 搜索二维矩阵 II | Search a 2D Matrix II_app

概述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 c

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：

[  [1,30]]

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 true13             } else if num < target {14                 r += 115             } else {16                 c -= 117             }18         }19         20         return false21     }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 false12     }13     14     var i = m-1,j = 015     16     while i >= 0,j < n {17         if matrix[i][j] == target {18             return true19         } else if matrix[i][j] > target {20             i -= 121         } else {22             j += 123         }24     }25     26     return false27     }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 -= 112             } else if matrix[row][column] < target {13                 row += 114             } else {15                 return true16             }17         }18 19         return false20     }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 -= 112             }13             if i < 0 || j >= n { return false }14         }15         return false16     }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 }