Leetcode Unique Paths II

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.

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解题思路：

Dynamic Programming.

解题思路与 Leetcode Unique Paths 一致。 dp[i][j] = dp[i-1][j] + dp[i][j-1] 

但有些方面需要修改。

1. 当(i, j)有障碍时dp[i][j] = 0

2. dp[0][j]和dp[i][0]未必为1.

　　dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]

　　dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]

3. 当obstacleGrid [0][0] = 1 或 obstacleGrid [m-1][n-1] = 1 时，return 0

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Java code:

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

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Reference:

1. http://www.cnblogs.com/springfor/p/3886644.html

2. http://bangbingsyb.blogspot.com/2014/11/leetcode-unique-paths-i-ii.html