给定一个包含非负整数的 m x n 网格 grid ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。
说明：每次只能向下或者向右移动一步示例 1：
1 3 1
1 5 1
4 2 1
输入：grid = [[1,3,1],[1,5,1],[4,2,1]]
输出：7
解释：因为路径 1→3→1→1→1 的总和最小。
示例 2：
输入：grid = [[1,2,3],[4,5,6]]
输出：12提示：
m == grid.length
n == grid[i].length
1 <= m, n <= 200
0 <= grid[i][j] <= 200

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解题方法：【动态规划】

二维数组：n * m
递推公式：
dp[i][j]表示从"Start"走到(i,j)位置的最小路径和
dp[i][j] = min(dp[i - 1][j] + grid[i][j], dp[i][j - 1] + grid[i][j])
初始化：
dp[0][0] = grid[0][0]
第一行元素的最小路径和，为从左到右元素和
for i in range(1, n):dp[0][i] = dp[0][i - 1] + grid[0][i]
第一列元素的最小路径和，为从上到下元素和
for j in range(1, m):dp[j][0] = dp[j - 1][0] + grid[j][0]

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class Solution:def different_roads(self, graph):#   行数row_number = len(nums)#   列数column_number = len(graph[0]) if graph[0] else 0  #   默认输入每行的列数都相同for row in graph:if not row:continuecolumn_number = max(column_number, len(row))    #   每行的列数不同，统计最大的列数dp = [[0]* column_number for _ in range(row_number)]dp[0][0] = graph[0][0]#   初始化首行for i in range(1, column_number):dp[0][i] = dp[0][i -1] + graph[0][i]#   初始化首列for j in range(row_number):dp[j][0] = dp[j - 1][0] + graph[j][0]#   递推公式for i in range(1, row_number):for j in range(1, column_number):dp[i][j] = min(dp[i - 1][j] + graph[i][j], dp[i][j - 1] + graph[i][j])return dp[row_number - 1][column_number - 1]if __name__ == '__main__':try:nums = eval(input())solution = Solution()result = solution.different_roads(nums)print(result)except Exception as e:print(e)

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仅作为代码记录，方便自学自查自纠