来自北大算法课的Leetcode题解：64. 最小路径和

代码仓库：​​Github | Leetcode solutions @doubleZ0108 from Peking University.​​

二维爬楼梯问题

• 解法1(T83% S43%): 经典动态规划

​​dp[i][j]​​​：从​​(0,0)​​​走到​​(i,j)​​​的路径长
目标：​​dp[m][n]​​转移方程：​​dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]​​初始化：第0行和第0列就是​​(0,0)​​直着走到自己

• 改进(T83% S81%): 直接用原数组进行动态规划，不需要开辟新空间$O(1)$

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        # 改进
        m, n = len(grid), len(grid[0])

        for i in range(1, m):
            grid[i][0] = grid[i][0] + grid[i-1][0]
        for j in range(1, n):
            grid[0][j] = grid[0][j] + grid[0][j-1]

        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[m-1][n-1]
    
    def otherSolution(self, grid):
        # 解法1
        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] = grid[i][0] + dp[i-1][0]
        for j in range(1, n):
            dp[0][j] = grid[0][j] + dp[0][j-1]


        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]

        return dp[m-1][n-1]