63. Unique Paths II -Medium

Question

*Follow up for “Unique Paths”:

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Note: m and n will be at most 100.

跟进”Unique Paths”，现在在表格中加一个障碍物（标记为1的是障碍物），有多少种路径可以到达终点。（m和n不会超过100）

Example

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
[0,0,0],
[0,1,0],
[0,0,0]
]

The total number of unique paths is 2.

Solution

• 动态规划解。原理和原题一样，取左边和上面的方法之和即可，对于障碍物，我们只需把它设为dp[i] = 0即可，这样不会影响到后面的计算。额外要考虑的是起始点和终点会不会有障碍物（居然有这样的测试集。。）。

  class Solution(object):
      def uniquePathsWithObstacles(self, obstacleGrid):
          """
          :type obstacleGrid: List[List[int]]
          :rtype: int
          """
          if len(obstacleGrid) == 0 or len(obstacleGrid[0]) == 0: return 0
          # 如果起点或终点为1，则不可能有路径
          if obstacleGrid[0][0] == 1 or obstacleGrid[-1][-1] == 1: return 0
  
          dp = [[0 for _ in range(len(obstacleGrid[0]))] for _ in range(len(obstacleGrid))]
          dp[0][0] = 1
          for index_r in range(len(obstacleGrid)):
              for index_c in range(len(obstacleGrid[0])):
                  if index_c == 0 and index_r == 0: continue
                  # 如果该表格是障碍，跳过
                  if obstacleGrid[index_r][index_c] == 1: continue
                  # 第一行的来源只能为左边（障碍物表格的dp为0，所以其右边也都是0）
                  if index_r == 0:
                      dp[index_r][index_c] = dp[index_r][index_c - 1]
                  # 第一行的来源只能为上面（障碍物表格的dp为0，所以其下边也都是0）
                  elif index_c == 0:
                      dp[index_r][index_c] = dp[index_r - 1][index_c]
                  # 其他都是来自上面或左边（因为障碍的dp为0，所以它不影响结果）
                  else:
                      dp[index_r][index_c] = dp[index_r - 1][index_c] + dp[index_r][index_c - 1]
          return dp[-1][-1]