dfs

idf · idf · commit 91653d89e8a7 · 2016-09-30T11:14:04.000-07:00
diff --git a/329 Longest Increasing Path in a Matrix.py b/329 Longest Increasing Path in a Matrix.py
@@ -0,0 +1,74 @@
+"""
+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.
+"""
+__author__ = 'Daniel'
+
+
+class Solution(object):
+    def __init__(self):
+        self.cache = None
+        self.dirs = ((-1, 0), (1, 0), (0, -1), (0, 1),)
+
+    def longestIncreasingPath(self, matrix):
+        """
+        dfs + cache
+        :type matrix: List[List[int]]
+        :rtype: int
+        """
+        if not matrix: return 0
+
+        m, n = len(matrix), len(matrix[0])
+        self.cache = [[None for _ in xrange(n)] for _ in xrange(m)]
+        gmax = 1
+        for i in xrange(m):
+            for j in xrange(n):
+                gmax = max(gmax, self.longest(matrix, i, j))
+
+        return gmax
+
+    def longest(self, matrix, i, j):
+        """
+        Strictly increasing, thus no need to have a visited matrix
+        """
+        if not self.cache[i][j]:
+            m, n = len(matrix), len(matrix[0])
+            maxa = 1
+            for d in self.dirs:
+                I, J = i + d[0], j + d[1]
+                if 0 <= I < m and 0 <= J < n and matrix[I][J] > matrix[i][j]:
+                    maxa = max(maxa, 1 + self.longest(matrix, I, J))
+
+            self.cache[i][j] = maxa
+
+        return self.cache[i][j]
+
+
+if __name__ == "__main__":
+    assert Solution().longestIncreasingPath([
+        [9, 9, 4],
+        [6, 6, 8],
+        [2, 1, 1]
+    ]) == 4
