思路

记忆化深搜

记录每个点的最长距离：

1. 如果搜索到该点发现该点没有值，则该点记录为1
2. 如果搜索到该点发现有值，则该值为搜索到点的最长路径，则在该值基础上加1即可
   顺序遍历

代码

public class Solution {
    
    // 备忘录
    int[][] memo;

    public int longestIncreasingPath(int[][] matrix) {
    
        if (null == matrix || matrix.length == 0
                || null == matrix[0] || matrix[0].length == 0) {
    
            return 0;
        }
        memo = new int[matrix.length][matrix[0].length];
        int res = 1;

        for (int i = 0; i < matrix.length; i++) {
    
            for (int j = 0; j < matrix[i].length; j++) {
    
                res = Math.max(dfs(matrix, i, j, Integer.MIN_VALUE), res);
            }
        }

        return res;
    }

    // 深搜
    public int dfs(int[][] matrix, int i, int j, int val) {
    
        if (i < 0 || i >= matrix.length || j < 0 || j >= matrix[i].length) {
    
            return 0;
        }
        if (matrix[i][j] <= val) {
    
            return 0;
        }
        if (memo[i][j] != 0) {
    
            return memo[i][j];
        }

        // 能走到这，先初始化1
        memo[i][j] = 1;

        int top = dfs(matrix, i + 1, j, matrix[i][j]);
        int bottom = dfs(matrix, i - 1, j, matrix[i][j]);
        int right = dfs(matrix, i, j + 1, matrix[i][j]);
        int left = dfs(matrix, i, j - 1, matrix[i][j]);

        // 记忆该点最长距离
        memo[i][j] = Math.max(Math.max(top, bottom), Math.max(right, left)) + 1;

        return memo[i][j];
    }
}