minimum falling path sum medium

Author: weiy

Array/MinimumFallingPathSum.py

@@ -0,0 +1,52 @@
+"""
+Given a square array of integers A, we want the minimum sum of a falling path through A.
+
+A falling path starts at any element in the first row, and chooses one element from each row.  The next row's choice must be in a column that is different from the previous row's column by at most one.
+
+ 
+
+Example 1:
+
+Input: [[1,2,3],[4,5,6],[7,8,9]]
+Output: 12
+Explanation: 
+The possible falling paths are:
+[1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
+[2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
+[3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
+The falling path with the smallest sum is [1,4,7], so the answer is 12.
+
+ 
+
+Note:
+
+1. 1 <= A.length == A[0].length <= 100
+2. -100 <= A[i][j] <= 100
+
+
+
+从上到下，每下层元素可以选择位于它之上的 左中右 三个元素。
+
+直接 Dp 思路：
+从第二层开始，每个元素都选择位于它之上的三个元素中最小的一个元素。
+最后输出最后一层中最小的元素即可。
+
+测试地址：
+https://leetcode.com/contest/weekly-contest-108/problems/minimum-falling-path-sum/
+
+"""
+class Solution(object):
+    def minFallingPathSum(self, A):
+        """
+        :type A: List[List[int]]
+        :rtype: int
+        """
+        
+        for y in range(1, len(A)):
+            for x in range(len(A[0])):
+                a = A[y-1][x-1] if x-1 >= 0 else float('inf')
+                b = A[y-1][x+1] if x+1 < len(A[0]) else float('inf')
+                
+                A[y][x] += min(A[y-1][x], a, b)
+        
+        return min(A[-1])