学步园

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.

思路：

这时候不能使用Unique Path中用到的排列组合技巧了，似乎必须做图搜索了？

其实给定左上角的点的Path值为1（假设这个点没有被放障碍物的话），那么到达其他任何一个点的可能路径有：

if blocked
 0
else
 path_from_left + path_from_top

也就是说，只要不被放上障碍物，那么可能路径为左边方格的值和上面方格的值的和，否则是0。据此可以填表，表格不需要重新申请内存，直接覆盖原来的网格数据即可。

题解：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        const size_t RR = obstacleGrid.size();
        const size_t CC = obstacleGrid[0].size();
        
        // fill up the grid system
        for(size_t r = 0; r < RR; ++r)
            for(size_t c = 0; c < CC; ++c)
            {
                if (obstacleGrid[r][c] == 1)
                    // blocked...
                    obstacleGrid[r][c] = 0;
                else if (r == 0 && c == 0)
                    // starting point
                    obstacleGrid[r][c] = 1;
                else {
                    int right_ways = (c == 0 ? 0 : obstacleGrid[r][c-1]);
                    int top_ways = (r == 0 ? 0 : obstacleGrid[r-1][c]);
                    obstacleGrid[r][c] = right_ways + top_ways;
                }
            }
        
        return obstacleGrid[RR - 1][CC - 1];
    }
};