题目：
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
代码：oj测试通过 Runtime: 102 ms
1 class Solution: 2 # @param grid, a list of lists of integers 3 # @return an integer 4 def minPathSum(self, grid): 5 # none case 6 if grid is None: 7 return None 8 # store each step on matrix 9 step_matrix=[[0 for i in range(len(grid[0]))] for i in range(len(grid))] 10 # first row 11 tmp_sum = 0 12 for i in range(len(grid[0])): 13 tmp_sum = tmp_sum + grid[0][i] 14 step_matrix[0][i] = tmp_sum 15 # first line 16 tmp_sum = 0 17 for i in range(len(grid)): 18 tmp_sum = tmp_sum + grid[i][0] 19 step_matrix[i][0] = tmp_sum 20 # dynamic program other positions in gird 21 for row in range(1,len(grid)): 22 for col in range(1,len(grid[0])): 23 step_matrix[row][col]=grid[row][col]+min(step_matrix[row-1][col],step_matrix[row][col-1]) 24 # return the minimun sum of path 25 return step_matrix[len(grid)-1][len(grid[0])-1]
思路：
动态规划常规思路。详情见http://www.cnblogs.com/xbf9xbf/p/4250359.html这道题。
需要注意的是，动态规划迭代条件step_matrix[][]=grid[][]+min(...)
min里面的一定要是step_matrix对应位置的元素。一开始写成grid对应的元素，第一次没有AC，修改后第二次AC了。