LeetCode 63. Unique Paths II Java

题目：

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.

题意：给出一个二维格子，其中值为1的点表示障碍点，要求求出从最左上角的点到最右下角的点有多少种走法。使用动态规划，对于其中一个点obstacleGrid[i][j](1<i<m,i<j<n)，到该点的走法为d(obstacleGrid[i][j])=d(obstacleGrid[i-1][j])+d(obstacleGrid[i][j-1])，对于第一行和第一列，如果该点前面有障碍点，那么到到此点有0中方法，反之为1。遍历数组即可求解。

代码：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int row = obstacleGrid.length;
        if(row ==0) return 0;
        int col = obstacleGrid[0].length;
        int[][] res = new int[row][col];
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(obstacleGrid[i][j]==1) {
                    res[i][j] = 0;
                    continue;
                }
                if(i==0 || j==0){
                    if(j!=0){
                        res[i][j] =  res[i][j-1];
                    }else if(i!=0){
                        res[i][j] =  res[i-1][j];
                    }else{
                        res[i][j] =  1;
                    }
                }else{
                    res[i][j] = res[i-1][j] + res[i][j-1];
                }
            }
        }
        return res[row-1][col-1];
    }
}