63. Unique Paths II java solutions

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.

该题和

Unique Paths java solutions

相比就是多了个障碍设置，只要dp初始化对障碍特殊设置一下，在中间dp 的过程加下判断即可。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        boolean flag = true;
        for(int i = 0; i < m; i++){
            if(obstacleGrid[i][0] == 0 && flag == true){
                dp[i][0] = 1;
            }else{
                flag = false;
                dp[i][0] = 0;
            }
        }
        flag = true;
        for(int i = 0; i < n; i++){
            if(obstacleGrid[0][i] == 0 && flag == true){
                dp[0][i] = 1;
            }else{
                flag = false;
                dp[0][i] = 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];
    }
}

 优化为一维数组解法：

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

            dp[j]  上一行，

dp[j-1]           当前位置的左边

 

因此可以优化为一维数组。