63. Unique Paths II(有障碍的路径 动态规划)

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Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

 class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
         int rows = obstacleGrid.length;
         int cols = obstacleGrid[0].length;
         int[][] dp = new int[rows][cols];

         for(int i = 0; i < rows;i++){
             for (int j = 0; j < cols ;j++){
                 if(obstacleGrid[i][j]==1)
                     dp[i][j] = 0;
                 else{
                     if (i==0&& j==0)
                         dp[i][j] = 1;
                     else if (i==0)
                         dp[i][j] = dp[i][j-1]; //边界 

                     else if (j == 0)
                         dp[i][j] = dp[i-1][j];
                     else
                         dp[i][j] = dp[i-1][j] + dp[i][j-1];
                 }
             }
         }
         return dp[rows-1][cols-1];
     }
 }

 class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
         int rows = obstacleGrid.length;
         int cols = obstacleGrid[0].length;
         for(int i = 0; i < rows;i++){
             for (int j = 0; j < cols ;j++){
                 if(obstacleGrid[i][j]==1)
                     obstacleGrid[i][j] = 0;
                 else if (i==0&& j==0)
                     obstacleGrid[i][j] = 1;
                 else if (i==0)
                     obstacleGrid[i][j] = obstacleGrid[i][j-1]*1; //边界，没有路径了，要么是0，要么是1

                 else if (j == 0)
                     obstacleGrid[i][j] = obstacleGrid[i-1][j]*1;
                 else
                     obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
             }
         }
         return obstacleGrid[rows-1][cols-1];
     }
 }