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LintCode : Unique Paths II

时间:2016-05-15 21:46:29      阅读:205      评论:0      收藏:0      [点我收藏+]

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Problem Description:

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.

Code:

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public class Solution {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // write your code here
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
            return 0;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] paths = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 0) {
                paths[i][0] = 1;
            } else {
                break;
            }
        }
        
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 0) {
                paths[0][i] = 1;
            } else {
                break;
            }
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 0) {
                    paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
                } else {
                    paths[i][j] = 0;
                }
            }
        }
        
        return paths[m - 1][n - 1];
    }
}
View Code

Hint:

Same as Unique Path, but you have to pay attention to situation where you can not get to there.

LintCode : Unique Paths II

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原文地址:http://www.cnblogs.com/dingjunnan/p/5496115.html

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