Unique Paths II

最新推荐文章于 2025-01-13 17:23:47 发布

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最新推荐文章于 2025-01-13 17:23:47 发布

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本文探讨了在网格中寻找不经过特定障碍物的独特路径的问题，提供了解决此类问题的算法，并通过实例展示了如何计算包含障碍物的3x3网格中的独特路径数量。

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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.

加入对矩阵元素是否为1的判断来初始化数组，只能使用二维数组。

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;
		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][j-1]+dp[i-1][j];  
            }  
        }
		return dp[m-1][n-1]; 
    }
}