[LeetCode 329] Longest Increasin

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： DFS + memorization

1. 如果只用backtracking会超时
2. 需要用一个记忆体matrix，对每个格子记录从它出发能达到的最长path length。
3. DFS返回就是当前这个格子能够得到的最长path length，再找到最大的那个，就是结果。
   • DFS中，如果当前格子已经有值了，就说明已经找到了从它出发能够找到的最大长度，直接返回结果
   • 再对其四个方向出发的邻居格子，递归调用DFS求解，currentCellPath = 1 + neighborCellPath
   • 最后返回 max currentCellPath
     转自http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-329-longest-increasing-path-in-a-matrix/

image.png

class Solution {
    /************** backtracking DFS TimeOut****************************
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        
        int[] longestPath = { 0 };
        boolean [][] visited = new boolean[matrix.length][matrix[0].length];
        
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                List<Integer> path = new ArrayList<> ();
                path.add (matrix[i][j]);
                visited[i][j] = true;
                longestIncreasingPathHelper (matrix, visited, longestPath, path, i, j);  //current path length
                visited[i][j] = false;
            }
        }
        
        return longestPath[0];
    }
    
    public void longestIncreasingPathHelper (int[][] matrix, 
                                             boolean[][] visited, 
                                             int[] longestPath, 
                                             List<Integer> path, 
                                             int row, int col) {
        boolean canContinue = row >= 0 && row < matrix.length && col >= 0 && col < matrix[0].length &&
                            (matrix[row][col] > path.get (path.size () - 1) || path.size () == 1);
        
        if (!canContinue) {
            return;
        }
        
        // path.add (matrix[row][col]);
        // visited[row][col] = true;
        // System.out.println (Arrays.toString (path.toArray()));
        longestPath[0] = Math.max (longestPath[0], path.size ());
        
        int[][] directions = {{-1, 0},{0, 1},{1, 0},{0, -1}};
        for (int[] direction : directions) {
            int nextRow = row + direction[0];
            int nextCol = col + direction[1];
            
            path.add (matrix[row][col]);
            visited[row][col] = true;
            // System.out.println (Arrays.toString (path.toArray()));
            
            
            longestIncreasingPathHelper (matrix, visited, longestPath, path, nextRow, nextCol);
            
            path.remove (path.size () - 1);
            visited[row][col] = false;
        }
 
    }
    *******************************************************/
    
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        
        int longestPath = 0;
        int[][] longestPathEachCell = new int[matrix.length][matrix[0].length];
        
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                longestPath = Math.max (longestPath, longestIncreasingPathHelper (matrix,longestPathEachCell, i, j));
            }
        }
        
        return longestPath;
    }
    
    // for each cell, return its max path
    public int longestIncreasingPathHelper (int[][] matrix, int[][] longestPathEachCell, int row, int col) {
        // if it already has, then directly return it. AVOID TOO MANY DUPLICATED EXECUTION
        if (longestPathEachCell[row][col] != 0)
            return longestPathEachCell[row][col];
        
        longestPathEachCell[row][col] = 1;
        int[][] directions = {{-1, 0},{0, 1},{1, 0},{0, -1}};
        
        for (int[] direction : directions) {
            int nextRow = row + direction[0];
            int nextCol = col + direction[1];
            
            boolean canContinue = nextRow >= 0 && nextRow < matrix.length && nextCol >= 0 && nextCol < matrix[0].length && matrix[nextRow][nextCol] > matrix[row][col];
            if (!canContinue)
                continue;
            
            longestPathEachCell[row][col] = Math.max (longestPathEachCell[row][col], 1 + longestIncreasingPathHelper(matrix, longestPathEachCell, nextRow, nextCol)); 
        }
        
        return longestPathEachCell[row][col];
    }
}