LeetCode64：Minimum Path Sum

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

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LeetCode：链接

动态规划的特点要求利用到上一次的结果，是一种特殊的迭代思想，动态规划的关键是要得到递推关系式。对于本题，从原点到达（i, j）的最小路径等于 ：原点到达（i-1, j）最小路径与到达（i, j-1）最小路径中的最小值，即 dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]。而且本题也可以不申请额外空间，直接在grid中修改参数即可，这使得空间复杂度得到了有效的利用，这类做法在实际场景中常常被用到。

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        m = len(grid)
        n = len(grid[0])
        dp = [[0] * n for i in range(m)]
        for i in range(m):
            for j in range(n):
                if i == 0 and j == 0:
                    dp[i][j] = grid[i][j]
                elif i == 0 and j > 0:
                    dp[i][j] = dp[i][j-1] + grid[i][j]
                elif j == 0 and i > 0:
                    dp[i][j] = dp[i-1][j] + grid[i][j]
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
        return dp[m-1][n-1]