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:

nums = [
  [9,9,4],
  [6,6,8],
  [2
 
,1,1]
]

Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return 4

The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

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這題第一思路是DFS，對於每個點出發都數longest increasing path，程式碼如下

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return 0;
        int res = Integer.MIN_VALUE;
        int[][] cache = new int[matrix.length][matrix[0].length];
        for(int i = 0; i < matrix.length; i++){
            for(int j = 0; j < matrix[0].length; j++){
                int max = helper(matrix, i, j, Integer.MIN_VALUE, cache);
                res = Math.max(res, max);
            }
        }
        
        return res;
    }
    
    private int helper(int[][] matrix, int i, int j, int val, int[][]cache){
        if(i < 0 || i >= matrix.length || j < 0 || j >= matrix[0].length)
            return 0;
        if(cache[i][j] != 0){
            return cache[i][j];
        }
        if(matrix[i][j] <= val){
            return 0;
        }
        int up = helper(matrix, i - 1, j, matrix[i][j], cache);
        int down = helper(matrix, i + 1, j, matrix[i][j], cache);
        int left = helper(matrix, i, j - 1, matrix[i][j], cache);
        int right = helper(matrix, i, j + 1, matrix[i][j], cache);
        
        return Math.max(Math.max(up, down), Math.max(left, right)) + 1;
    }
}
 

Complexity Analysis

• Time complexity :
  O(2^{m+n})
  
  The search is repeated for each valid increasing path. In the worst case we can have
  O(2^{m+n}) calls. For example:

1 23 . . . n

2 3. . .   n+1

3 . . .     n+2

.           .

.           .

.           .

m m+1. . . n+m-1

• Space complexity :
  O(mn). For each DFS we need O(h) space used by the system stack, whereh is the maximum depth of the recursion. In the worst case,  O(h) = O(mn)

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return 0;
        int res = Integer.MIN_VALUE;
        int[][] cache = new int[matrix.length][matrix[0].length];  
        for(int i = 0; i < matrix.length; i++){
            for(int j = 0; j < matrix[0].length; j++){
                int max = helper(matrix, i, j, Integer.MIN_VALUE, cache);
                res = Math.max(res, max);
            }
        }
        
        return res;
    }
    
    private int helper(int[][] matrix, int i, int j, int val, int[][]cache){
        if(i < 0 || i >= matrix.length || j < 0 || j >= matrix[0].length)
            return 0;
        if(matrix[i][j] <= val){
            return 0;
        }
        if(cache[i][j] != 0){
            return cache[i][j] + 1;
        }else{
            int up = helper(matrix, i - 1, j, matrix[i][j], cache);
            int down = helper(matrix, i + 1, j, matrix[i][j], cache);
            int left = helper(matrix, i, j - 1, matrix[i][j], cache);
            int right = helper(matrix, i, j + 1, matrix[i][j], cache);
            
            int max =  Math.max(Math.max(up, down), Math.max(left, right));
            cache[i][j] = max;
            return max + 1;
        }
    }
}