LeetCode-63.Unique Paths II

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

使用dp,要判断当前位置是不是1，若是1则表示当前位置值为0，否则使用递推式计算；

public int uniquePathsWithObstacles(int[][] obstacleGrid) {//my dp
        //应该做判空处理，这里没有判断
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] flag = new int[m+1][n+1]; 
        if(obstacleGrid[0][0]==1){//如果0 0 位置是1，则表示没有路径可达
            return 0;
        }
        for(int i = 1;i <= m;i++){
            for(int j = 1;j<=n;j++){
                if(obstacleGrid[i-1][j-1]==1){
                    flag[i][j] = 0;
                }
                else{
                    if(1 == i && 1==j){
                        flag[i][j] =1;
                    }
                    else{
                        flag[i][j] = flag[i-1][j]+flag[i][j-1];
                    }                    
                }
                
            }
        }
        return flag[m][n];
    }

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不同的路径  LeetCode62   https://www.cnblogs.com/zhacai/p/10941798.html