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
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]
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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