题解 | #矩阵的最小路径和#

矩阵的最小路径和

https://www.nowcoder.com/practice/7d21b6be4c6b429bb92d219341c4f8bb

#
# 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
#
# 
# @param matrix int整型二维数组 the matrix
# @return int整型
#
class Solution:
    def minPathSum(self , matrix: List[List[int]]) -> int:
        # write code here
        row = len(matrix)
        col = len(matrix[0])
        dp =[[0]*(col) for _ in range(row)]
        for i in range(row):
            for j in range(col):
                if i==0 and j==0:
                    dp[i][j]=matrix[i][j]
                elif i==0 and j!=0:
                    dp[i][j] =dp[i][j-1]+matrix[i][j]
                elif i!=0 and j==0:
                    dp[i][j] =dp[i-1][j]+matrix[i][j]
                else: 
                    dp[i][j]=min(dp[i-1][j],dp[i][j-1])+matrix[i][j]
        print(dp)
        return dp[-1][-1]

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