329. Longest Increasing Path in a Matrix

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.

---------------------------
Solution #1, typical DFS, 没啥好说的

class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0) return 0;
        int m = matrix.length, n = matrix[0].length;
        int[][] inter = new int[m][n];
        int longest = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                longest = Math.max(longest, dfs(matrix, i, j, inter, Integer.MAX_VALUE));
            }
        }
        
        return longest;
    }
    
    private int dfs(int[][] matrix, int row, int col, int[][] inter, int pre) {
        if (row < 0 || row >= matrix.length || col < 0 || col >= matrix[0].length 
           || matrix[row][col] >= pre) return 0;
        
        if (inter[row][col] > 0) return inter[row][col];
        
        int longest = dfs(matrix, row + 1, col, inter, matrix[row][col]);
        longest = Math.max(longest, dfs(matrix, row - 1, col, inter, matrix[row][col]));
        longest = Math.max(longest, dfs(matrix, row, col + 1, inter, matrix[row][col]));
        longest = Math.max(longest, dfs(matrix, row, col - 1, inter, matrix[row][col]));
        inter[row][col] = longest + 1;
        
        return longest + 1;
    }
}

Solution #2, iterative。也是典型的BFS。ToDo, 有空可以写一下
找到所有的起点，一起塞进Queue（或List），同时记录步数