63. Unique Paths II

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.

Note: m and n will be at most 100.

这道题是在unique path的基础上加上了障碍物，解决办法还是动态规划，代码如下：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        
        dp[m-1][n-1] = obstacleGrid[m-1][n-1]==1?0:1;
        for(int i=m-2;i>=0;i--){
            if(obstacleGrid[i][n-1]==1){
                dp[i][n-1]=0;
            }else{
                dp[i][n-1] =dp[i+1][n-1];
            }
        }
        for(int i=n-2;i>=0;i--){
            if(obstacleGrid[m-1][i]==1){
                dp[m-1][i] = 0;
            }else{
                dp[m-1][i] = dp[m-1][i+1];
            }
        }
        for(int i=m-2;i>=0;i--){
            for(int j=n-2;j>=0;j--){
                if(obstacleGrid[i][j]==1){
                    dp[i][j] =0;
                }else{
                    
                    dp[i][j] = dp[i][j+1]+dp[i+1][j];
                    
                }
            }
        }
        return dp[0][0];
    }
}

看了discussion后，发现大神的解法很简洁，用的办法我取名字为光的直线传播，代码如下：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int n = obstacleGrid[0].length;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i=0;i<obstacleGrid.length;i++){
            for(int j=0;j<obstacleGrid[i].length;j++){
                if(obstacleGrid[i][j]==1){
                    dp[j] = 0;
                }else if(j>0){
                    dp[j]+=dp[j-1];
                }
            }
        }
        return dp[n-1];
    }
}

这个解法也是用的dp方法解决的，只不过顺序是相反的，答案也是相反的，相当于镜面映射了。