Unique Paths II

　　原题链接：[https://leetcode.com/problems/unique-paths-ii/]。(https://leetcode.com/problems/unique-paths-ii/)

原题

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.
翻译：和上一题一样，不过多了一些障碍——矩阵中障碍为“1”，没有障碍的地方为“0”，还是问有多少种走法？

思路

　　思路还是DP（动态规划），区别就在于，如果矩阵的这个位置是“1”，那么这个地方的走法数就是0。
　　特别要注意的就只第一行和第一列的处理，如果没有遇见“1”，那么就是一种走法（和问题一一样），一旦遇到“1”，那么改点和之后的位置的走法数都是0了。

代码

　　问题比较简单，代码应该可以解释他自己。

int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m  = obstacleGrid.size();        if(m < 1) return 0;        int n = obstacleGrid[0].size();        if(n < 1) return 0;        if(m == 1 && n == 1) return 1 - obstacleGrid[0][0];        vector<vector<int>> theVector;        for(int i = 0; i < m; i++)        {            vector<int> temp(n);            theVector.push_back(temp);        }        bool flag = false;        for(int i = 0; i < n; i++)        {            if(flag || obstacleGrid[0][i] == 1)            {                theVector[0][i] = 0;                flag = true;            }            else                theVector[0][i] = 1;        }        flag = false;        for(int i = 0; i < m; i++)        {            if(flag || obstacleGrid[i][0] == 1)            {                theVector[i][0] = 0;                flag = true;            }            else                theVector[i][0] = 1;        }        for(int i = 1; i < m; i++)            for(int j = 1; j < n; j++)            {                if(obstacleGrid[i][j] == 1)                    theVector[i][j] = 0;                else                    theVector[i][j] = theVector[i - 1][j] + theVector[i][j - 1];            }        return theVector[m - 1][n - 1];    }

　　　接下来我们讨论讨论“Path sum”的问题。