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剑指 Offer II 112. 最长递增路径

题目描述

给定一个 m x n 整数矩阵 matrix ，找出其中 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。 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：matrix = [[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：matrix = [[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。

示例 3：

输入：matrix = [[1]]
输出：1

 

提示：

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 200
  • 0 <= matrix[i][j] <= 231 - 1

 

注意：本题与主站 329 题相同： https://leetcode.cn/problems/longest-increasing-path-in-a-matrix/

解法

方法一

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class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        @cache
        def dfs(i, j):
            ans = 1
            for a, b in [[-1, 0], [1, 0], [0, 1], [0, -1]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]:
                    ans = max(ans, dfs(x, y) + 1)
            return ans

        m, n = len(matrix), len(matrix[0])
        return max(dfs(i, j) for i in range(m) for j in range(n))

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class Solution {
    private int[][] memo;
    private int[][] matrix;
    private int m;
    private int n;

    public int longestIncreasingPath(int[][] matrix) {
        this.matrix = matrix;
        m = matrix.length;
        n = matrix[0].length;
        memo = new int[m][n];
        for (int i = 0; i < m; ++i) {
            Arrays.fill(memo[i], -1);
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (memo[i][j] != -1) {
            return memo[i][j];
        }
        int ans = 1;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                ans = Math.max(ans, dfs(x, y) + 1);
            }
        }
        memo[i][j] = ans;
        return ans;
    }
}

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class Solution {
public:
    vector<vector<int>> memo;
    vector<vector<int>> matrix;
    int m;
    int n;

    int longestIncreasingPath(vector<vector<int>>& matrix) {
        m = matrix.size();
        n = matrix[0].size();
        memo.resize(m, vector<int>(n, -1));
        this->matrix = matrix;
        int ans = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                ans = max(ans, dfs(i, j));
        return ans;
    }

    int dfs(int i, int j) {
        if (memo[i][j] != -1) return memo[i][j];
        int ans = 1;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j])
                ans = max(ans, dfs(x, y) + 1);
        }
        memo[i][j] = ans;
        return ans;
    }
};

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func longestIncreasingPath(matrix [][]int) int {
    m, n := len(matrix), len(matrix[0])
    memo := make([][]int, m)
    for i := range memo {
        memo[i] = make([]int, n)
        for j := range memo[i] {
            memo[i][j] = -1
        }
    }
    ans := -1
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if memo[i][j] != -1 {
            return memo[i][j]
        }
        ans := 1
        dirs := []int{-1, 0, 1, 0, -1}
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j] {
                ans = max(ans, dfs(x, y)+1)
            }
        }
        memo[i][j] = ans
        return ans
    }
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            ans = max(ans, dfs(i, j))
        }
    }
    return ans
}

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class Solution {
    private var memo: [[Int]] = []
    private var matrix: [[Int]] = []
    private var m: Int = 0
    private var n: Int = 0

    func longestIncreasingPath(_ matrix: [[Int]]) -> Int {
        self.matrix = matrix
        m = matrix.count
        n = matrix[0].count
        memo = Array(repeating: Array(repeating: -1, count: n), count: m)

        var ans = 0
        for i in 0..<m {
            for j in 0..<n {
                ans = max(ans, dfs(i, j))
            }
        }
        return ans
    }

    private func dfs(_ i: Int, _ j: Int) -> Int {
        if memo[i][j] != -1 {
            return memo[i][j]
        }
        var ans = 1
        let dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        for (dx, dy) in dirs {
            let x = i + dx, y = j + dy
            if x >= 0, x < m, y >= 0, y < n, matrix[x][y] > matrix[i][j] {
                ans = max(ans, dfs(x, y) + 1)
            }
        }
        memo[i][j] = ans
        return ans
    }
}

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