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

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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

class Solution {
    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
        // start at top right
        if matrix.count < 1 || matrix[0].count < 1 { return false }
        
        let rows = matrix.count, cols = matrix[0].count
        var r = 0, c = cols - 1
        
        while c >= 0, r < rows {
            let num = matrix[r][c]
            if num == target {
                return true
            } else if num < target {
                r += 1
            } else {
                c -= 1
            }
        }
        
        return false
    }
}

---

332ms

class Solution {
    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
    guard matrix.count > 0, matrix[0].count > 0 else {
        return false
    }
    
    let m = matrix.count
    let n = matrix[0].count
    
    if target < matrix[0][0] || target > matrix[m-1][n-1] {
        return false
    }
    
    var i = m-1, j = 0
    
    while i >= 0, j < n {
        if matrix[i][j] == target {
            return true
        } else if matrix[i][j] > target {
            i -= 1
        } else {
            j += 1
        }
    }
    
    return false
    }
}

---

332ms

class Solution {
    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
        guard !matrix.isEmpty && !matrix[0].isEmpty else {
            return false
        }

        var row = 0
        var column = matrix[0].count - 1
        while column >= 0 && row < matrix.count {
            if matrix[row][column] > target {
                column -= 1
            } else if matrix[row][column] < target {
                row += 1
            } else {
                return true
            }
        }

        return false
    }
}

---

340ms

class Solution {
    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {

        guard matrix.count > 0, matrix[0].count > 0 else { return false }
        var m = matrix.count, n = matrix[0].count
        var i = m-1, j = 0
        while true {
            if matrix[i][j] == target { return true }
            else if matrix[i][j] < target { j += 1 }
            else {
                i -= 1
            }
            if i < 0 || j >= n { return false }
        }
        return false
    }
}

---

348ms

class Solution {
    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
        for i in 0..<matrix.count {
            for j in stride(from: matrix[i].count, to: 0, by: -1) {
                    if matrix[i][j - 1] == target {
                        return true
                    }
            }
        }
        
        return false 
    }
}