[LeetCode]64、最小路径和

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。

思路解析

找到一条从左上角到右下角的路径，路径之和最小。
每个元素，向右走或者向下走。两条路，找到一条路径之和最小的一个。
cost（i, j）= grid[i][j] + min(cost(i+1, j), cost(i, j+1))
需要求的就是cost(0, 0)
申请动态规划数组dp[m][n],初始化右下角的值grid[m-1][n-1]。考虑bottom和right，能够直接初始化。dp[m-1][j] = grid[i][j] + dp[i][j+1]
dp[i][n-1] = grid[i][n-1] + dp[i+1][n-1]
除此之外，就是状态转移dp[i][j] = grid[i][j] + min(dp[i+1][j], dp[i][j+1])

class Solution:
    def minPathSum(self, grid):
        if not grid:
            return 0
        m = len(grid)
        n = len(grid[0])
        dp = [[0 for _ in range(n)] for _ in range(m)]
       
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                if i == m-1 and j != n-1:
                    dp[i][j] = grid[i][j] + dp[i][j+1]
                elif j == n-1 and i != m-1:
                    dp[i][j] = grid[i][j] + dp[i+1][j]
                elif i != m-1 and j != n-1:
                    dp[i][j] = grid[i][j] + min(dp[i][j+1], dp[i+1][j])
                else:
                    dp[i][j] = grid[i][j]
        return dp[0][0]

AC64

优化

每一个状态依赖右侧或者下侧。利用最后一行，没有向下的，动态转移赋值。在上个解法中，我们可以用一个一维数组来代替二维数组，dp 数组的大小和列大小相同。这是因为对于某个固定状态，只需要考虑下方和右侧的节点。首先初始化 dp 数组最后一个元素是右下角的元素值，然后我们向左移更新每个 dp(j) 为：

dp(j)=grid(i,j)+min(dp(j),dp(j+1))

我们对于每一行都重复这个过程，然后向上一行移动，计算完成后 dp(0) 就是最后的结果。

class Solution:
#     def minPathSum(self, grid):
#         if not grid:
#             return 0
#         m = len(grid)
#         n = len(grid[0])
#         dp = [[0 for _ in range(n)] for _ in range(m)]
       
#         for i in range(m-1, -1, -1):
#             for j in range(n-1, -1, -1):
#                 if i == m-1 and j != n-1:
#                     dp[i][j] = grid[i][j] + dp[i][j+1]
#                 elif j == n-1 and i != m-1:
#                     dp[i][j] = grid[i][j] + dp[i+1][j]
#                 elif i != m-1 and j != n-1:
#                     dp[i][j] = grid[i][j] + min(dp[i][j+1], dp[i+1][j])
#                 else:
#                     dp[i][j] = grid[i][j]
#         return dp[0][0]
    def minPathSum(self, grid):
        if not grid:
            return 0
        m = len(grid)
        n = len(grid[0])
        dp = [0 for _ in range(n)]

        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if i == m - 1 and j != n - 1:
                    dp[j] = grid[i][j] + dp[j + 1]
                elif j == n - 1 and i != m - 1:
                    dp[j] = grid[i][j] + dp[j]
                elif i != m - 1 and j != n - 1:
                    dp[j] = grid[i][j] + min(dp[j + 1], dp[j])
                else:
                    dp[j] = grid[i][j]
        return dp[0]

image.png

优化

直接在数组上修改，减少空间复杂度

最后编辑于：2019.08.04 09:31:08