Leetcode unique PathII

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.
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原文链接：https://blog.youkuaiyun.com/chaochen1407/article/details/43307579

package com.company;

public class LeetCodeUniquePathII {

   /* public int uniquePath2(int a[][]){
        int x=a.length;
        int y=a[0].length;

        int nets[][] =new int[x][y];

        //row
        for(int i =0;i<x;i++){
            for(int j =0;j<y;j++){
                nets[i][j]=0;
                if(a[i][j]==1){
                    nets[i][j]=0;

                }

                if(nets[i-1][j]=0&&i=0){
                    nets[i][j]=0;
                } else if(nets[i][j-1]=0&&j=0){
                    nets[i][j]=0;
                }else if(nets[i-1][j]=0){

                    nets[i][j]=nets[i-1][j];
                }else if (nets[i][j-1]=0){
                    nets[i][j]=nets[i][j-1];
                }else{
                    nets[i][j]=nets[i-1][j]+nets[i][j-1];
                }
            }
        }
        return nets[i-1][j-1];
    }*/


    public int uniquePath3(int a[][]) {

        int x = a.length;
        int y = a[0].length;

        int nets[][] = new int[x][y];

        //row
        for (int i = 0; i < x; i++) {
            for (int j = 0; j < y; j++) {
                nets[i][j] = 0;
                if (a[i][j] == 1) {
                    nets[i][j] = 0;
                } else {
                    if (i == 0 && j == 0) {
                        nets[i][j] = 1;
                    } else {
                        if (i - 1 >= 0) {
                            nets[i][j] += nets[i - 1][j];
                        }
                        if (j - 1 >= 0) {
                            nets[i][j] += nets[1][j - 1];
                        }

                    }
                }
            }
        }
        return nets[x - 1][y - 1];
    }


}