# 931. Minimum Falling Path Sum 下降路径最小和

@TOC

## # 题目描述

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`

## # 解题方法

### # 动态规划

``````class Solution(object):
def minFallingPathSum(self, A):
"""
:type A: List[List[int]]
:rtype: int
"""
M, N = len(A), len(A[0])
dp = [[0] * (N + 2) for _ in range(M)]
for i in range(M):
dp[i][0] = dp[i][-1] = float('inf')
for j in range(1, N + 1):
dp[i][j] = A[i][j - 1]
for i in range(1, M):
for j in range(1, N + 1):
dp[i][j] = A[i][j - 1] + min(dp[i - 1][j - 1], dp[i - 1][j], dp[i - 1][j + 1])
return min(dp[-1])
``````

## # 参考资料

https://leetcode.com/problems/minimum-falling-path-sum/discuss/186689/Java-DP-solution-with-graph-illustrated-explanations

## # 日期

2018 年 10 月 28 日 —— 啊，悲伤的周赛