329. Longest Increasing Path in a Matrix

最新推荐文章于 2019-04-13 11:46:37 发布

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最新推荐文章于 2019-04-13 11:46:37 发布

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文章标签： java python

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10101569.html

本文探讨了在二维矩阵中寻找最长递增路径的问题，提供了三种实现方式：Java的DFS+DP，C++的DFS+DP，以及Python的DFS+DP。通过递归和记忆化搜索，文章详细解释了如何在不移动对角线的情况下找到从任一单元格开始的最长递增路径。

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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))

转载于:https://www.cnblogs.com/ruruozhenhao/p/10101569.html