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.
与Unique Path, Minimum Path Sum题目解法类似，都是DP问题。建立一个新的矩阵，矩阵每个元素记录的是从upper left到该点的路径数。与Unique Path不同的是：写递归式的时候，能不能代入下一层递归需要检验obstacleGrid里该点是不是一个obstacle，如果是，该层返回0且不继续递归
一维DP解法：
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int[][] grid = obstacleGrid; 4 int xLen = grid.length; 5 if (xLen==0) return 0; 6 int yLen = grid[0].length; 7 if (yLen==0) return 0; 8 9 int[] path = new int[yLen]; 10 if (grid[0][0]==1) 11 return 0; 12 else path[0] = 1; 13 for (int i=1;i<yLen;i++) 14 if (grid[0][i]==1) path[i]=0; 15 else path[i] = path[i-1]; 16 17 for (int i=1;i<xLen;i++){ 18 if (grid[i][0]==1) path[0] = 0; 19 for (int j=1;j<yLen;j++) 20 if (grid[i][j]==1) path[j] = 0; 21 else path[j] = path[j-1]+path[j]; 22 } 23 return path[yLen-1]; 24 25 } 26 }
Iteration方法：21-26行可删，因为如果那里是obstacle，那里在res矩阵里也是0，加它也无妨
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int m = obstacleGrid.length; 4 int n = obstacleGrid[0].length; 5 int[][] res = new int[m][n]; 6 for (int k=0; k<n; k++) { 7 if (obstacleGrid[0][k] != 1) { 8 res[0][k] = 1; 9 } 10 else break; 11 } 12 for (int t=0; t<m; t++) { 13 if (obstacleGrid[t][0] != 1) { 14 res[t][0] = 1; 15 } 16 else break; 17 } 18 for (int i=1; i<m; i++) { 19 for (int j=1; j<n; j++) { 20 if (obstacleGrid[i][j] == 1) continue; 21 else { 22 res[i][j] = res[i-1][j] + res[i][j-1]; 23 } 24 } 25 } 26 return res[m-1][n-1]; 27 } 28 }