leetcode------Unique Paths II

标题：	Unique Paths II
通过率：	28%
难度：	中等

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.

与第一个版本一样，但是原始数组中会出现1的情况那么针对1来处理，如果遇到1则将其置-1，然后在路径相加中如果是-1则不进行相加

代码如下：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m=obstacleGrid.length;
        int n=obstacleGrid[0].length;
        boolean flag=true;
        if(obstacleGrid[0][0]==1)return 0;
        for(int i=0;i<m;i++){
            if(obstacleGrid[i][0]!=1&&flag)
                obstacleGrid[i][0]=1;
            else {
                obstacleGrid[i][0]=-1;
                flag=false;
            }
        }
        flag=true;
        for(int i=1;i<n;i++){
            if(obstacleGrid[0][i]!=1&&flag)
                obstacleGrid[0][i]=1;
            else {
                obstacleGrid[0][i]=-1;
                flag=false;
            }
        }
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                if(obstacleGrid[i][j]!=1){
                    if(obstacleGrid[i-1][j]==-1&&obstacleGrid[i][j-1]==-1)
                        obstacleGrid[i][j]=0;
                    if(obstacleGrid[i-1][j]==-1)
                        obstacleGrid[i][j]+=obstacleGrid[i][j-1];
                    else if(obstacleGrid[i][j-1]==-1)
                        obstacleGrid[i][j]+=obstacleGrid[i-1][j];
                    else obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
                }
                else{
                    obstacleGrid[i][j]=-1;
                }
            }
        }
            if(obstacleGrid[m-1][n-1]==-1)return 0;
            else 
                return obstacleGrid[m-1][n-1];
        
    }
}