[leetcode][63] Unique Paths II

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

解析

同Unique Path，中间加上石头，加上额外判断即可。

参考答案（自己写的）

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int[][] matrix = new int[obstacleGrid.length][obstacleGrid[0].length];
        for (int i = 0; i< matrix.length; i++) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            matrix[i][0] = 1;
        }
        
        for (int i = 0; i < matrix[0].length; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            matrix[0][i] = 1;
        }
        
        for (int i = 1; i < matrix.length; i++) {
            for (int j = 1; j < matrix[0].length; j++) {
                if (obstacleGrid[i][j] != 1) {
                    matrix[i][j] = matrix[i-1][j] + matrix[i][j-1];
                }
            }
        }
        
        return matrix[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
    }
}

基本思路和Unique Path一致，注意几个点：

• 如果某个点有石头，那到达这点路径数等于0；
• 第0行的某个点有石头，那么次点右边的的点路径数都为0；
• 第0列的某个点有石头，那么次点下边的的点路径数都为0；