代码随想录算法训练营day42 | 62.不同路径、63. 不同路径 II

62.不同路径

• 确定dp数组及其下标的含义：dp[i][j]代表机器人到达i行j列时不同的路径数
• 确定递推公式：dp[i][j] = dp[i-1][j] + dp[i][j-1]
• dp数组的初始化：dp[0][j] = 1 dp[i][0] = 1
• 确定遍历顺序：从前向后
• 举例推导dp数组

  class Solution:
      def uniquePaths(self, m: int, n: int) -> int:
          dp = [[1] * n for _ in range(m)]
          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[-1][-1]

  可以使用一维数组优化空间复杂度

  
  （图片来源网络，侵删）

  class Solution:
      def uniquePaths(self, m: int, n: int) -> int:
          dp = [1] * n
          for i in range(1, m):
              for j in range(1, n):
                  dp[j] += dp[j-1]
          return dp[-1]

  63. 不同路径 II

  • 确定dp数组以及下标的含义：dp[i][j]代表到达第i行第j列的不同路径数
  • 确定递推公式：if obstacleGrid[i][j] == 1: dp[i][j] = 0 else: dp[i][j] = dp[i-1][j] + dp[i][j-1]
  • dp数组的初始化：第0行和第0列都需要初始化为1，遇到障碍物后后续都初始化为0
  • 确定遍历顺序：从前向后遍历
  • 举例推导dp数组

    class Solution:
        def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
            m = len(obstacleGrid)
            n = len(obstacleGrid[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:
                        dp[i][j] = 0
                    else:
                        dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            return dp[-1][-1]