一、题目描述

给定一个包含非负整数的 m x n 网格，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：

输入:
[
  [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。

二、代码实现

方法一：二维状态转移方程
dp[i][j] = Min( dp[i - 1][j] ,dp[i][j - 1] ) + grid[i][j]
空间复杂度 O(m x n)

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

方法二：状态压缩，一维状态转移方程
上面的方法中，时间复杂度为 O(nm), 但是辅助空间也为 O(nm)。通过压缩空间的方法来减小辅助空间，使得辅助空间只需要 O(m)。
状态转移方程：dp[ i ] = min ( dp[ i - 1] , dp[ i ] ) + grid[ i ] [ j ]

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