题解 | #矩阵的最小路径和#

> - `dp[i][j]`表示矩阵中第i行第j列的最小路径和.

> - `dp[i][0]=dp[i-1][0]+matrix[i][0]` 并且 `dp[0][j]=dp[0][j-1]+matrix[0][j]`

> - `dp[i][j]=min(dp[i-1][j],dp[i][j-1])+matrix[i][j]`

> - `dp[-1][-1]` 即为最终答案

class Solution:
    def minPathSum(self , matrix: List[List[int]]) -> int:
        m = len(matrix)
        n = len(matrix[0])
        dp = [[0]*n for _ in range(m)]
        dp[0][0] = matrix[0][0]
        for i in range(1,m):
            dp[i][0] = dp[i-1][0] + matrix[i][0]
        for j in range(1,n):
            dp[0][j] = dp[0][j-1] + matrix[0][j]
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + matrix[i][j]
        return dp[-1][-1]