动态规划（2）

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-path-sum
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution:
    def minPathSum(self, grid: [[int]]) -> int:
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if i == j == 0: continue
                elif i == 0:  grid[i][j] = grid[i][j - 1] + grid[i][j]
                elif j == 0:  grid[i][j] = grid[i - 1][j] + grid[i][j]
                else: grid[i][j] = min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]
        return grid[-1][-1]

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """

        # 非优化动态规划，时间复杂度O(m*n)，空间复杂度O(n*m)
        m = len(grid)
        n = len(grid[0])
        for i in range(1,n):
            grid[0][i] += grid[0][i-1]
        for j in range(1,m):
            grid[j][0] += grid[j-1][0]
        for i in range(1,m):
            for j in range(1,n):
                grid[i][j] = min(grid[i-1][j],grid[i][j-1]) + grid[i][j]
        return grid[-1][-1]
        

        # 优化算法,即求每次到达位置的最优解
        # 定义状态：即定义数组元素的含义,dp[i]表示当前位置的最小数值综合
        # 建立状态转移方程:dp[i] = min(dp[i-1],dp[i])+grid[i,j]
        # 设定初始值:这一步是关键
        # 选择结果，即dp[-1]
        m = len(grid)
        n = len(grid[0])
        dp = [0]*n
        dp[0] = grid[0][0]
        for k in range(1,n):
            dp[k] = dp[k-1] + grid[0][k] 
        for i in range(1,m):
            for j in range(0,n):
                if j == 0 : dp[j] += grid[i][j]
                else:
                    dp[j] = min(dp[j-1],dp[j])+grid[i][j]
        return dp[-1]

作者：yu-fa-tang-you-dian-tian
链接：https://leetcode-cn.com/problems/minimum-path-sum/solution/python-dong-tai-gui-hua-you-hua-kong-jian-fu-za-du/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。