Leetcode: 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.

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：

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

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

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] res = new int[m][n];
        for (int k=0; k<n; k++) {
            if (obstacleGrid[0][k] != 1) {
                res[0][k] = 1;
            }
            else break;
        }
        for (int t=0; t<m; t++) {
            if (obstacleGrid[t][0] != 1) {
                res[t][0] = 1;
            }
            else break;
        }
        for (int i=1; i<m; i++) {
            for (int j=1; j<n; j++) {
                if (obstacleGrid[i][j] == 1) continue;
                else {
                    res[i][j] = res[i-1][j] + res[i][j-1];
                }
            }
        }
        return res[m-1][n-1];
    }
}