62. Unique Paths i & ii

62. Unique Paths

A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below).

How many possible unique paths are there?

DP  注意边界

public class Solution {
    public int uniquePaths(int m, int n) {
        int dp[][] = new int[n][m];
        for(int i = 0 ; i < n ; i++){
          dp[i][0] = 1;
        }
        for(int j = 0 ; j < m ; j ++){
          dp[0][j] = 1;
        }
        for(int i = 1; i < n ; i++){
            for(int j = 1 ; j < m ; j++){
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[n-1][m-1];
    }
}

63. 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.

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

        for(int i = 0 ; i < obstacleGrid[0].length; i++){
            if(obstacleGrid[0][i] != 1)
                dp[0][i] = 1;
            else
               break;
        }

        for(int i = 1; i < obstacleGrid.length ; i++){
            for(int j = 1; j < obstacleGrid[0].length ; j++){
                if(obstacleGrid[i][j] == 1)
                    dp[i][j] = 0;
                if(obstacleGrid[i][j] == 0)
                   dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }

        return dp[obstacleGrid.length-1][obstacleGrid[0].length -1];
    }
}