leetcode 63 不同的路径2

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<vector<int>>& obstacleGrid) {
    int m = obstacleGrid.size();
    int n = obstacleGrid[].size();

    for (int i = ; i < m; ++i) {
        for (int j = ; j < n; ++j) {
            if (obstacleGrid[i][j] == ) {
                obstacleGrid[i][j] = -;
                continue;
            }

            if (i ==  && j == )
                obstacleGrid[][] = ;
            else if (i == )
                obstacleGrid[][j] = obstacleGrid[][j - ];
            else if (j == )
                obstacleGrid[i][] = obstacleGrid[i - ][];
            else if (obstacleGrid[i - ][j] == - && obstacleGrid[i][j - ] == -)
                obstacleGrid[i][j] = -;
            else if (obstacleGrid[i - ][j] == -)
                obstacleGrid[i][j] = obstacleGrid[i][j - ];
            else if (obstacleGrid[i][j - ] == -)
                obstacleGrid[i][j] = obstacleGrid[i - ][j];
            else if (obstacleGrid[i][j] == )
                obstacleGrid[i][j] = -;
            else
                obstacleGrid[i][j] = obstacleGrid[i - ][j] + obstacleGrid[i][j - ];
        }
    }

    if (obstacleGrid[m - ][n - ] == -)
        return ;
    return obstacleGrid[m - ][n - ];
}