描述：

从左上角走到右下角，中间可能有若干阻碍；

题目给出一个矩阵，0表示可以走，1表示有障碍。

 

解决：

思路同第一题，只是如果上面或左边有障碍，自身不一定能走，注意些边界条件即可，复杂度仍是m*n。

为了防止和真正的路径1冲突，走过的障碍改为-1。

int uniquePathsWithObstacles(vector
     
      
       >& obstacleGrid) {    int m = obstacleGrid.size();    int n = obstacleGrid[0].size();        for (int i = 0; i < m; ++i) {        for (int j = 0; j < n; ++j) {            if (obstacleGrid[i][j] == 1) {                obstacleGrid[i][j] = -1;                continue;            }                        if (i == 0 && j == 0)                obstacleGrid[0][0] = 1;            else if (i == 0)                obstacleGrid[0][j] = obstacleGrid[0][j - 1];            else if (j == 0)                obstacleGrid[i][0] = obstacleGrid[i - 1][0];            else if (obstacleGrid[i - 1][j] == -1 && obstacleGrid[i][j - 1] == -1)                obstacleGrid[i][j] = -1;            else if (obstacleGrid[i - 1][j] == -1)                obstacleGrid[i][j] = obstacleGrid[i][j - 1];            else if (obstacleGrid[i][j - 1] == -1)                obstacleGrid[i][j] = obstacleGrid[i - 1][j];            else if (obstacleGrid[i][j] == 1)                obstacleGrid[i][j] = -1;            else                obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];        }    }        if (obstacleGrid[m - 1][n - 1] == -1)        return 0;    return obstacleGrid[m - 1][n - 1];}