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

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int r = obstacleGrid.length;
        int c = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1) return 0;
        obstacleGrid[0][0] = 1;
        for(int i = 1; i < r; i++){
            obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i-1][0] == 1) ? 1 : 0;//第一行只能从左到右，初始化第一格是1，后面要是遇到本来就有的1就暂停让其变0，这样后面就加不进去了
        }
        for(int i = 1; i < c; i++){
            obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i-1] == 1) ? 1 : 0;
        }
        for(int i = 1; i < r; i++){
            for(int j = 1; j < c; j++){
                if(obstacleGrid[i][j] == 0){
                    obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
                }
                else{
                    obstacleGrid[i][j] = 0;//遇到原始的1（obstacle）让其置0
                }
            }
        }
        return obstacleGrid[r-1][c-1];
    }
}

每一次只能走一格，所以ij有两种可能，从【i-1】【j】和从【i】【j-1】，每一步存着到这一步的步数，所以用dp。

Hint1：The robot can only move either down or right. Hence any cell in the first row can only be reached from the cell left to it. However, if any cell has an obstacle, you don't let that cell contribute to any path. So, for the first row, the number of ways will simply be

if obstacleGrid[i][j] is not an obstacle
     obstacleGrid[i,j] = obstacleGrid[i,j - 1] 
else
     obstacleGrid[i,j] = 0

You can do a similar processing for finding out the number of ways of reaching the cells in the first column.

Hint2：For any other cell, we can find out the number of ways of reaching it, by making use of the number of ways of reaching the cell directly above it and the cell to the left of it in the grid. This is because these are the only two directions from which the robot can come to the current cell.

Hint3：Since we are making use of pre-computed values along the iteration, this becomes a dynamic programming problem.

if obstacleGrid[i][j] is not an obstacle
     obstacleGrid[i,j] = obstacleGrid[i,j - 1]  + obstacleGrid[i - 1][j]
else
     obstacleGrid[i,j] = 0