Longest Increasing Path in a Matrix (H)

题目

Given an m x n integers matrix, return the length of the longest increasing path in matrix.

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

Example 1:

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

Example 2:

Input: matrix = [[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.

Example 3:

Input: matrix = [[1]]
Output: 1

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• 0 <= matrix[i][j] <= 2^31 - 1

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题意

给定一个矩阵，可以从任一位置出发得到一条递增的路径，求最长路径的长度。

思路

DFS+记忆化很容易解决。

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代码实现

Java

class Solution {
    private int[] xShift = {0, -1, 0, 1};
    private int[] yShift = {-1, 0, 1, 0};
    private int m, n;

    public int longestIncreasingPath(int[][] matrix) {
        m = matrix.length;
        n = matrix[0].length;

        int ans = 0;
        int[][] record = new int[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                ans = Math.max(ans, dfs(matrix, i, j, record));
            }
        }

        return ans;
    }

    private int dfs(int[][] matrix, int x, int y, int[][] record) {
        if (record[x][y] > 0) return record[x][y];

        record[x][y] = 1;
        for (int i = 0; i < 4; i++) {
            int nx = x + xShift[i], ny = y + yShift[i];
            if (nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[x][y] < matrix[nx][ny]) {
                record[x][y] = Math.max(record[x][y], 1 + dfs(matrix, nx, ny, record));
            }
        }
        return record[x][y];
    }
}