[8.2] Robot in a Grid

Imagine a robot sitting on the upper left corner of grid with r rows and c columns. The robot can only move in two directions, right and down, but certain cells are 'off limit' such that the robot cannot step on them. Design an algorithm to find a path for the robot from the top left to the bottom right.

Similar questions in Leetcode:

https://leetcode.com/problems/unique-paths/

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] paths = new int[m][n];
        for(int i = 0; i < m; ++i) {
            for(int j = 0; j < n; ++j) {
                if(i == 0 && j == 0) {
                    paths[0][0] = 1;
                } else {
                    paths[i][j] = (i==0 ? 0: paths[i - 1][j]) + (j ==0 ? 0: paths[i][j - 1]);
                }
            }
        }
        return paths[m - 1][n - 1];
    }
}

https://leetcode.com/problems/unique-paths-ii/

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

 

posted @ 2016-12-08 10:39  Phoebe815  阅读(364)  评论(0编辑  收藏  举报