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：
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int m = obstacleGrid.length; 4 int n = obstacleGrid[0].length; 5 int[][] dp = new int[m][n]; 6 7 dp[m-1][n-1] = obstacleGrid[m-1][n-1]==1?0:1; 8 for(int i=m-2;i>=0;i--){ 9 if(obstacleGrid[i][n-1]==1){ 10 dp[i][n-1]=0; 11 }else{ 12 dp[i][n-1] =dp[i+1][n-1]; 13 } 14 } 15 for(int i=n-2;i>=0;i--){ 16 if(obstacleGrid[m-1][i]==1){ 17 dp[m-1][i] = 0; 18 }else{ 19 dp[m-1][i] = dp[m-1][i+1]; 20 } 21 } 22 for(int i=m-2;i>=0;i--){ 23 for(int j=n-2;j>=0;j--){ 24 if(obstacleGrid[i][j]==1){ 25 dp[i][j] =0; 26 }else{ 27 28 dp[i][j] = dp[i][j+1]+dp[i+1][j]; 29 30 } 31 } 32 } 33 return dp[0][0]; 34 } 35 }
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：
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int n = obstacleGrid[0].length; 4 int[] dp = new int[n]; 5 dp[0] = 1; 6 for(int i=0;i<obstacleGrid.length;i++){ 7 for(int j=0;j<obstacleGrid[i].length;j++){ 8 if(obstacleGrid[i][j]==1){ 9 dp[j] = 0; 10 }else if(j>0){ 11 dp[j]+=dp[j-1]; 12 } 13 } 14 } 15 return dp[n-1]; 16 } 17 }
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。
原文：http://www.cnblogs.com/codeskiller/p/6385683.html