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Unique Paths II

时间:2015-06-25 02:14:19      阅读:250      评论:0      收藏:0      [点我收藏+]

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 m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if (m <= 0 || n <= 0) {
        	return 0;
        }
        if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1)  
            return 0; 
        int[][] res = new int[m][n];
        res[0][0] = 1;
        for (int i = 1; i < m; i++) {
        	if (obstacleGrid[i][0] == 1) {
        		res[i][0] = 0;
        	} else {
        		res[i][0] = res[i-1][0];
        	}
        }
        for (int i = 1; i < n; i++) {
        	if (obstacleGrid[0][i] == 1) {
        		res[0][i] = 0;
        	} else {
        		res[0][i] = res[0][i-1];
        	}
        }
        for (int i = 1; i < m; i++) {
        	for (int j = 1; j < n; j++) {
        		if (obstacleGrid[i][j] == 1) {
        			res[i][j] = 0;
        		} else {
        			res[i][j] = res[i-1][j] + res[i][j-1];
        		}
        	}
        }
        return res[m-1][n-1];
    }
}

?

Unique Paths II

原文:http://hcx2013.iteye.com/blog/2221569

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