【DP】64. Minimum Path Sum

问题描述：

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.

解题思路：

很明显使用动态规划求解。创建 dp[m][n]，其中 dp[i][j] 表示位置 (i, j) 的最小累加和。最后 dp[-1][-1] 就是答案。

转移方程也很简单：
dp[i][j] = grid[i][j] + min(dp[i][j-1], dp[i-1][j])

注意：编程时，需要初始化第一行和第一列。后一个数由前面的数累加得到的，即：

• 左上角元素：dp[0][0] = grid[0][0]
• 第一列初始化：dp[i][0] = dp[i-1][0] + grid[i][0]
• 第一行初始化：dp[0][j] = dp[0][j-1] + grid[0][j]

Python3 实现：

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = grid[0][0]
        for i in range(1, m):
            dp[i][0] = dp[i-1][0] + grid[i][0]
        for j in range(1, n):
            dp[0][j] = dp[0][j-1] + grid[0][j]
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = grid[i][j] + min(dp[i][j-1], dp[i-1][j])
        return dp[-1][-1]

print(Solution().minPathSum([[1,3,1],[1,5,1],[4,2,1]]))  # 7 (1->3->1->1->1)

转载于:https://www.jianshu.com/p/32e752fd6b8f