Unique Paths II

本文探讨了在网格中加入障碍物后,独特的路径数量如何计算。通过实例展示了解决此类问题的方法。

Question:

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.

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int row = obstacleGrid.length;
        int column = obstacleGrid[0].length;
        int[][] totalPath = new int[row][column];
        if (obstacleGrid[0][0] == 1) {
            totalPath[0][0] = 0;
        } else {
        	totalPath[0][0] = 1;
        }
        
        for (int i = 1; i < row; i++) {
        	if (obstacleGrid[i][0] == 1) {
        		totalPath[i][0] = 0;
        	} else {
        		totalPath[i][0] = totalPath[i - 1][0];
        	}
        }
        for (int j = 1; j < column; j++) {
        	if (obstacleGrid[0][j] == 1) {
        		totalPath[0][j] = 0;
        	} else {
        		totalPath[0][j] = totalPath[0][j - 1];
        	}
        }
        
        for (int i = 1; i < row; i++) {
        	for (int j = 1; j < column; j++) {
        		if (obstacleGrid[i][j] == 1) {
        			totalPath[i][j] = 0;
        		} else {
        			totalPath[i][j] = totalPath[i - 1][j] + totalPath[i][j - 1];
        		}
        	}
        }
        return totalPath[row-1][column-1];           
    }
}


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