329. Longest Increasing Path in a Matrix

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.

 

Approach #1 Java: [dfs + dp]

class Solution {
    public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0) return 0;
        int m = matrix.length, n = matrix[0].length;
        int[][] cache = new int[m][n];
        int max = 1;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int len = dfs(matrix, i, j, m, n, cache);
                max = Math.max(max, len);
            }
        }
        return max;
    }
    
    public int dfs(int[][] matrix, int i, int j, int m, int n, int[][] cache) {
        if (cache[i][j] != 0) return cache[i][j];
        int max = 1;
        for (int[] dir : dirs) {
            int x = i + dir[0], y = j + dir[1];
            if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j]) continue;
            int len = 1 + dfs(matrix, x, y, m, n, cache);
            max = Math.max(max, len);
        }
        cache[i][j] = max;
        return max;
    }
}

　　

Approach #2: C++.

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if (matrix.size() == 0) return 0;
        int row = matrix.size();
        int col = matrix[0].size();
        vector<vector<int>> memory(row, vector<int>(col, 0));
        
        std::function<int(int, int)> dfs = [&] (int x, int y) {
            if (memory[x][y]) return memory[x][y];
            for (auto &dir : dirs) {
                int xx = x + dir.first;
                int yy = y + dir.second;
                if (xx < 0 || xx >= row || yy < 0 || yy >= col || matrix[x][y] >= matrix[xx][yy]) continue;
                memory[x][y] = std::max(memory[x][y], dfs(xx, yy));
            }
            return ++memory[x][y];
        };
        
        int ans = 0;
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                ans = max(ans, dfs(i, j));
            }
        }
        return ans;
    }
private:
    vector<pair<int, int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
};

　　

Approach #3: Python.

class Solution(object):
    def longestIncreasingPath(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: int
        """
        def dfs(i, j):
            if not dp[i][j]:
                val = matrix[i][j]
                dp[i][j] = 1 + max(
                    dfs(i-1, j) if i and val > matrix[i-1][j] else 0,
                    dfs(i+1, j) if i < M-1 and val > matrix[i+1][j] else 0,
                    dfs(i, j-1) if j and val > matrix[i][j-1] else 0,
                    dfs(i, j+1) if j < N-1 and val > matrix[i][j+1] else 0
                )
            return dp[i][j]
            
        if not matrix or not matrix[0]: return 0
        M, N = len(matrix), len(matrix[0])
        dp = [[0] * N for i in range(M)]
        return max(dfs(x, y) for x in range(M) for y in range(N))