题解 | #矩阵的最小路径和#

#
# 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
        n = len(matrix) # 矩阵的行数
        m = len(matrix[0]) # 矩阵的列数

        if n==1 and m==1: # 特殊情况，1x1矩阵直接放好
            return matrix[0][0]

        #dp[i][j]表示从matrix[0][0]到matrix[i][j]最短路径和
        #初始化dp[i][j]值为0
        dp = [[0]*n for _ in range(m)]
        
        dp[0][0] = matrix[0][0]
        for i in range(1,n):
            dp[i][0] = matrix[i][0] + dp[i-1][0]
        for i in range(1,m):
            dp[0][i] = matrix[0][i] + dp[0][i-1]
                
        for i in range(1,n):
            for j in range(1,m):
                if dp[i-1][j]>dp[i][j-1]:
                    dp[i][j] = matrix[i][j] + dp[i][j-1]
                else:
                    dp[i][j] = matrix[i][j] + dp[i-1][j]

        return dp[n-1][m-1]