leetcode - Unique Paths II

探讨了在网格中加入障碍物后,唯一路径数量的变化。通过动态规划的方法解决了问题,并给出了具体实例。

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.

For 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.

Note: m and n will be at most 100.

class Solution {
public:
    int uniquePathsWithObstacles(std::vector<std::vector<int> > &obstacleGrid) {
        std::vector<std::vector<int>> dp(obstacleGrid.size(),std::vector<int>(obstacleGrid[0].size(),0));
		dp[0][0] = obstacleGrid[0][0] ? 0 : 1;
		for (int i = 1; i < obstacleGrid.size(); i++)
		{
			dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0];
		}
		for (int i = 1; i < obstacleGrid[0].size(); i++)
		{
			dp[0][i] = obstacleGrid[0][i] ? 0 : dp[0][i-1];
		}
		for (int i = 1; i < obstacleGrid.size(); i++)
		{
			for (int j = 1; j < obstacleGrid[0].size(); j++)
			{
				dp[i][j] = obstacleGrid[i][j] ? 0 : dp[i-1][j] + dp[i][j-1];
			}
		}
		return dp[obstacleGrid.size()-1][obstacleGrid[0].size()-1];
    }
};


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