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 <= A.length == A[0].length <= 100
-100 <= A[i][j] <= 100
动态规划, 初始化dp为A[0], 从A[1]开始遍历, 状态转移方程:
dp = [val + min(dp[max(0,c-1)], dp[c], dp[min(col-1, c+1)]) for c, val in enumerate(A[r])]
最后返回min(dp)即可.
class Solution(object):
def minFallingPathSum(self, A):
"""
:type A: List[List[int]]
:rtype: int
"""
dp = A[0]
col = len(A[0])
for r in range(1, len(A)):
dp = [val + min(dp[max(0,c-1)], dp[c], dp[min(col-1, c+1)]) for c, val in enumerate(A[r])]
return min(dp)