63. Unique Paths II

Description:

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.

简要题解：

采用动态规划。

将obstacleGrid抽象为一个有向图，每个值为0格子i对应图中的一个结点i。如果obstacleGrid中可以从格子i移动到格子j，则在对应的有向图中存在一条从结点i到结点j的边。

子问题c：令c(i, j)为以obstacleGrid[0][0]为起点，obstacleGrid[i][j]为终点的路径个数（为了方便起见，当i或j小于0时，c(i, j) = 0）。

第一个子问题：当obstacleGrid[0][0]不为1时，c(0, 0) = 1；否则，c(0, 0) = 0。

c(i, j)与c(i - 1, j)和c(i, j - 1)的关系：c(i, j) = c(i - 1, j) + c(i, j - 1)

从第一个子问题开始不断迭代求解下一个子问题，直到所有问题都得到解决。就自然能得到所要的结果。

代码：

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