【LeetCode】(63)Unique Paths II(Medium)

题目

Unique Paths II

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Question  Solution 

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.

解析

其实就是前面一个题目的改变，不过是增加了是否是路障的判断

需要注意的是，可能起点和终点就是路障。同时我们只用了一维数组来存储，而每次并不会求第一列的path值，因为默认为1，这里需要进行判断。代码如下

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
          int m = obstacleGrid.size();
		  int n = obstacleGrid[0].size();

		  vector<int> path(n);
		  path[0] = 1 - obstacleGrid[0][0];

		  for (int i = 0;i<m;i++)
		  {
			  if (obstacleGrid[i][0] == 1)
			  {
				  path[0] = 0;
			  }
			  for (int j = 1;j<n ;j++)
			  {
				  if (obstacleGrid[i][j] == 0)
				  {
					  path[j] +=  path[j-1];
				  }
				  else
					  path[j] = 0;
			  }
		  }
		  return path[n-1];
    }
};

大神代码如下

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) 
{
    const int m = obstacleGrid.size();
    const int n = obstacleGrid[0].size();
    if (obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) 
        return 0;
    vector<int> f(n, 0);
    f[0] = obstacleGrid[0][0] ? 0 : 1;
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            f[j] = obstacleGrid[i][j] ? 0 : (j == 0 ? 0 : f[j - 1]) + f[j];
    return f[n - 1];
}
};