dp[i][j] 表示从(0 ,0)出发到(i, j) 有dp[i][j]的路径数
边界值处理，(0 ,0)到(i,0)或(0, j)，只有一条路径
对于当前位置，只能由左方和上方位置到达，故有状态转移方程：dp[i][j] = dp[i-1][j] + dp[i][j-1]

func uniquePaths(m int, n int) int {dp := make([][]int, m)for i, _ := range dp {dp[i] = make([]int, n)}for i := 0; i < m; i++ {for j := 0; j < n; j++ {if i*j == 0 {dp[i][j] = 1} else {dp[i][j] = dp[i-1][j] + dp[i][j-1]}}}return dp[m-1][n-1]
}

若起点或者终点为障碍物，直接返回0；
边界值处理，(0 ,0)到(i,0)或(0, j)，若不遇到障碍物，只有1条路径，若遇到障碍物，到达当前位置及之后位置路径数为0。
若遇到障碍物，到达当前位置(i, j)的路径数dp[i][j]=0，否则dp[i][j] = dp[i-1][j] + dp[i][j-1]

func uniquePathsWithObstacles(obstacleGrid [][]int) int {m, n := len(obstacleGrid), len(obstacleGrid[0])dp := make([][]int, m)if obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1 {return 0}for i, _ := range dp {dp[i] = make([]int, n)}for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {dp[i][0] = 1}for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {dp[0][j] = 1}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]}}}return dp[m-1][n-1]
}

空间优化：滑动数组