[LeetCode][Java] Unique Paths II

题目：

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.

题意：

紧跟着题目《Unique Paths》，现给出这样一题目：

假设在格子中加入一些障碍，会出现多少存在且唯一的不同路径呢？

障碍和空白格子分别被标记为1 and 0 .

比方一个3x3的格子中的中间存在一个障碍，例如以下所看到的：

[[0,0,0],[0,1,0],[0,0,0]
]

总的路径数为2.

算法分析：

     思路与题目《Unique Paths》类似，不同之处为：

     初始化边界上行和列时，出现障碍。后面路径数dp的都是0

     中间的格子出现障碍时，该格子dp表示的路径数直接填0

AC代码：

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