[Swift]LeetCode329. 矩阵中的最长递增路径 | Longest Increasing Path in a Matrix

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/10260831.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！！！

Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:

Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

---

给定一个整数矩阵，找出最长递增路径的长度。

对于每个单元格，你可以往上，下，左，右四个方向移动。 你不能在对角线方向上移动或移动到边界外（即不允许环绕）。

示例 1:

输入: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
输出: 4 
解释: 最长递增路径为 [1, 2, 6, 9]。

示例 2:

输入: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
输出: 4 
解释: 最长递增路径是 [3, 4, 5, 6]。注意不允许在对角线方向上移动。

---

 284 ms

class Solution {
   func longestIncreasingPath(_ matrix: [[Int]]) -> Int {
        
        if matrix.isEmpty || matrix[0].isEmpty {
            return 0
        }
        
        let m = matrix.count
        let n = matrix[0].count
        
        var counts = Array(repeating: Array(repeating: 0, count: n), count: m)
        var longest = 0
        
        for i in 0..<m {
            for j in 0..<n {
                let c = longestHelp(matrix, &counts, i, j)
                longest = max(longest, c)
            }
        }
        
        return longest
    }
    
    func longestHelp(_ matrix: [[Int]], _ counts : inout [[Int]], _ x : Int , _ y : Int) ->Int  {
        if counts[x][y] != 0 {
            return counts[x][y]
        }
        
        var res = 1
        if x > 0 && matrix[x-1][y] > matrix[x][y] {
            res = max(res, longestHelp(matrix, &counts, x-1, y) + 1)
        }
        if x < matrix.count-1 && matrix[x+1][y] > matrix[x][y] {
            res = max(res, longestHelp(matrix, &counts, x+1, y) + 1)
        }
        
        if y > 0 && matrix[x][y-1] > matrix[x][y] {
            res = max(res, longestHelp(matrix, &counts, x, y-1) + 1)
        }
        if y < matrix[0].count-1 && matrix[x][y+1] > matrix[x][y] {
            res = max(res, longestHelp(matrix, &counts, x, y+1) + 1)
        }
        
        counts[x][y] = res
        return res
    }
}

---

404ms

class Solution {
    var dp: [[Int]] = []
    let dirs: [[Int]] = [[-1, 0], [0, -1], [1, 0], [0, 1]]
    func longestIncreasingPath(_ matrix: [[Int]]) -> Int {
        if matrix.count == 0 { return 0 }
        dp = [[Int]](repeating: [Int](repeating: -1, count: matrix[0].count), count: matrix.count)
        var res = 0
        for r in 0..<matrix.count {
            for c in 0..<matrix[0].count {
                res = max(res, dfs(matrix, r, c))
            }
        }
        
        return res
    }
    
    private func dfs(_ matrix: [[Int]], _ r: Int, _ c: Int) -> Int {
        if self.dp[r][c] != -1 {
            return dp[r][c]
        }
        
        var res = 0
        for dir in dirs {
            let newr  = r+dir[0], newc =  c + dir[1]
            if newr >= 0 && newc >= 0 &&
                newr < matrix.count && newc < matrix[0].count &&
                matrix[newr][newc] > matrix[r][c] {
                    res = max(res, dfs(matrix, newr, newc))
                }
            
        }
        dp[r][c] = res+1
        return res + 1
    }
}