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]