题解 | #矩阵的最小路径和#
矩阵的最小路径和
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# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
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# @param matrix int整型二维数组 the matrix
# @return int整型
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class Solution:
def minPathSum(self , matrix: List[List[int]]) -> int:
# write code here
n =len(matrix)
m =len(matrix[0])
dp =[[1 for j in range(n)] for i in range(m)]
for i in range(n):
for j in range(m):
if i==0 and j==0:
dp[i][j]=matrix[0][0]
elif i==0 :
dp[i][j]=dp[i][j-1]+matrix[i][j]
elif j==0 :
dp[i][j]=dp[i-1][j]+matrix[i][j]
elif 0<i<m and 0<j <n:
dp[i][j]=min(dp[i-1][j],dp[i][j-1])+matrix[i][j]
return dp[n-1][m-1]

