力扣-931.下降路径最小和

由于是从第一行任意一个位置开始下降，所以先初始化第一行的值

每个元素有三种选择可以到达

最左边的元素可以从[i-1][j]，[i-1][j+1] 即上方，右上方获得

最右边的元素可以从[i-1][j-1]，[i-1][j] 即左上方，上方获得

中间元素可以从[i-1][j-1]，[i-1][j] ，[i-1][j+1]即左上方，上方，右上方获得

class Solution(object):
    def minFallingPathSum(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: int
        """
        m = len(matrix)
        n = len(matrix[0])
        dp = [[999999 for i in range(n)] for j in range(m)]

        for j in range(0, n):
            dp[0][j] = matrix[0][j]

        for i in range(1, m):
            for j in range(n):
                if j == 0:
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j + 1]) + matrix[i][j]
                elif j == n - 1:
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j]) + matrix[i][j]
                else:
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i - 1][j + 1]) + matrix[i][j]

        res = min(dp[m - 1])
        return res


if __name__ == '__main__':
    matrix = [[-19, 57], [-40, -5]]
    Sol = Solution()
    res = Solution.minFallingPathSum(Sol, matrix)
    print(res)