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.

ref:

水中的鱼: [LeetCode] Unique Paths II 解题报告

[解题思路]

和Unique Path一样的转移方程：

Step[i][j] = Step[i-1][j] + Step[i][j-1] if Array[i][j] ==0
or            = 0 if Array[i][j] =1

 

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

 

 

滚动数组

class Solution {

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int row = obstacleGrid.length;
        if(row == 0) return 0; 
        int col = obstacleGrid[0].length;
        
        if(obstacleGrid[0][0] == 1)
            return 0;
        
        int[] paths = new int[col];
        paths[0] = 1; 
        for(int i = 0; i < row ; i++){
            for(int j = 0; j < col; j++){
                if(obstacleGrid[i][j] == 1){
                    paths[j] = 0;
                }else if(j > 0){
                    paths[j] = paths[j]+paths[j-1];
                }
            }
        }
        return paths[col -1];
    }
}

 

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