題目描述：

方法一：動態規劃 O(mnlogmn)

class Solution(object):
    def longestIncreasingPath(self, matrix):
        if not matrix or not matrix[0]:
            return 0
        m, n = len(matrix), len(matrix[0])
        lst = []
        for i in range(m):
            for j in range(n):
                lst.append((matrix[i][j], i, j))
        lst.sort()
        dp 
 
= [[0 for _ in range(n)] for _ in range(m)]
        for num, i, j in lst:
            dp[i][j] = 1
            for di, dj in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
                r, c = i + di, j + dj
                if 0 <= r < m and 0 <= c < n:
                    if matrix[i][j] > matrix[r][c]:
                        dp[i][j] 
 
= max(dp[i][j], 1 + dp[r][c])
        return max([dp[i][j] for i in range(m) for j in range(n)])

方法二：記憶化深搜 O(mn) O(mn)

class Solution(object):
    def longestIncreasingPath(self, matrix):
        if not matrix or not matrix[0]:
            return 0
        dirs = [(0,-1),(0,1),(1,0),(-1,0)]

        rows,columns 
 
= len(matrix),len(matrix[0])

        @lru_cache(None)
        def helper(row,column):
            best = 1
            for _i,_j in dirs:
                newrow,newcolumn = row + _i,column + _j
                if 0 <= newrow < rows and 0 <= newcolumn < columns and matrix[row][column] < matrix[newrow][newcolumn]:
                    best = max(best,helper(newrow,newcolumn) + 1)
            return best
        
        ans = 0
        for i in range(rows):
            for j in range(columns):
                ans = max(ans,helper(i,j))
        return ans