代码随想训练营第三十九天（Python）｜ 62.不同路径、63. 不同路径 II、343. 整数拆分

62.不同路径

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        # dp[i][j] 代表到达 dp[i][j] 有多少不同路径
        dp = [[0]*n for _ in range(m)]

        # 初始化
        for i in range(m):
            dp[i][0] = 1

        for j in range(n):
            dp[0][j] = 1

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

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

63. 不同路径 II

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        if obstacleGrid[0][0] == 1 or obstacleGrid[m-1][n-1] == 1:
            return 0
        dp = [[0]*n for _ in range(m)]

        for i in range(m):
            if obstacleGrid[i][0] == 0:
                dp[i][0] = 1
            else:
                break

        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break

        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                else:
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]

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

343. 整数拆分

class Solution:
    def integerBreak(self, n: int) -> int:
        # dp[i] 代表拆分整数 i 获取的最大值
        dp = [0] * (n+1)

        # 初始化 dp 数组
        dp[2] = 1

        # 遍历 dp 数组
        for i in range(2, n+1):
            # 从 1 开始拆分整数
            for j in range(1, i//2 + 1):
                dp[i] = max(dp[i], j*(i-j), j*dp[i-j])

        return dp[n]