[数据结构与算法]leetcode:不同路径2

思路：
在不同路径的基础上
1.初始化时若碰到障碍物，dp后面全部是0
2.在遍历求解时遇到障碍，dp[i][j] = 0

代码：

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
        //初始化
    m := len(obstacleGrid)
    n := len(obstacleGrid[0])
    dp := make([][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([]int, n)
    }
    for i := 0; i < n; i++ {
        if obstacleGrid[0][i] != 1{
            dp[0][i] = 1
        } else {
            break
        }
        
    }
    for i := 0; i < m; i++ {
        if obstacleGrid[i][0] != 1{
            dp[i][0] = 1
        } else {
            break
        }
        
    }

    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if obstacleGrid[i][j] != 1 {
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
            } else {
                dp[i][j] = 0
            } 
        }
    }

    return dp[m-1][n-1]

}