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.
给定一个整数A的正方形数组，我们想要通过A的下降路径的最小和。
下降路径从第一行中的任何元素开始，并从每行中选择一个元素。 下一行的选择必须位于与前一行的列不同的列中，最多一行。
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
python实践
class Solution:
def minFallingPathSum(self, A):
# 思路： 每一行先排列，然后选择首元素，即是最小的falling path,
# 此方法有问题
total = 0
for i in range(0, len(A)):
A[i] = sorted(A[i])
print(A[i])
total += A[i][0]
return total
# 此方法行不通
- 动态规划做法，参考
https://blog.csdn.net/fuxuemingzhu/article/details/83479398
#!/usr/bin/env python
# _*_ coding:utf-8 _*_
class Solution:
def minFallingPathSum(self, A):
# 思路： 每一行先排列，然后选择首元素，即是最小的falling path,
# 此方法有问题
total = 0
for i in range(0, len(A)):
A[i] = sorted(A[i])
print(A[i])
total += A[i][0]
return total
def minFallingPathSum2(self,A):
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])
# dp:[[inf, 17, 82, inf], [inf, 18, -27, inf]]
return min(dp[-1])
if __name__ == '__main__':
A = [[17,82],[1,-44]]
t = Solution().minFallingPathSum2(A)
print(t)
