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
Accepted
5,997
Submissions
10,729
Solution1:(TLE)
class Solution:
def minFallingPathSum(self, A):
"""
:type A: List[List[int]]
:rtype: int
"""
res = 99999999
l = len(A)
def solve(i,j,sum):
nonlocal res
if i==l-1:
# print(sum)
res = min(res,sum)
else:
solve(i+1,j,sum+A[i+1][j])
if j>0:
solve(i+1,j-1,sum+A[i+1][j-1])
if j<l-1:
solve(i+1,j+1,sum+A[i+1][j+1])
for i in range(l):
solve(0,i,A[0][i])
return res
36 / 46 test cases passed.
这样求出每条到最后一行的路径的方法时间复杂度较大，在每一行时，只使用到这个点的最小路径。
Solution2:
class Solution:
def minFallingPathSum(self, A):
"""
:type A: List[List[int]]
:rtype: int
"""
dp = [[0 for i in range(len(A))] for j in range(len(A))]
for i in range(len(A)):
for j in range(len(A)):
if i==0:
dp[i][j] = A[i][j]
continue
temp = A[i][j] + dp[i-1][j]
if j>0:
temp = min(temp,A[i][j] + dp[i-1][j-1])
if j<len(A)-1:
temp = min(A[i][j] + dp[i - 1][j+1],temp)
dp[i][j] = temp
return min(dp[-1][j] for j in range(len(A)))