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[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];
}