63.不同路径Ⅱ

比62多个判断即可

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        // 一条横线时，如果有障碍，那么就没有所需的路径
        for(int i = 0; i < m &&  obstacleGrid[i][0] == 0; i++){
            dp[i][0] = 1;
        }
        // 同上
        for(int i = 0; i < n &&  obstacleGrid[0][i] == 0; i++){
            dp[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                // 没遇到了障碍才可继续前进
                if (obstacleGrid[i][j] == 0) 
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}