115 · 不同的路径 II Unique Paths II

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这篇文章讨论了如何在含有障碍物的网格中计算独特的路径数量，通过动态规划的方法解决二维空间中的路径问题，给出一个名为Solution的Java代码实现。重点在于处理障碍元素并保持路径计数的准确性。

115 · 不同的路径 II Unique Paths II

描述
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.
Example 2:

Input:

obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output:

2
Explanation:

Only 2 different path

public class Solution {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // write your code here
        int m = obstacleGrid.length , n = obstacleGrid[0].length ;
        int[][] f = new int[m][n] ;
        if(obstacleGrid[0][0] == 1){
            return 0 ;
        }
        f[0][0] = 1 ;
        for(int i = 0 ; i < m ; i++){
            for(int j = 0 ; j < n ; j++){
                if(obstacleGrid[i][j] == 1){
                    f[i][j] = 0 ;
                }else{
                    if(j>0 && i == 0 ){
                        f[i][j] = f[i][j-1] ;
                    }else if(i>0 && j == 0){
                        f[i][j] = f[i-1][j] ;
                    }else if (i > 0 && j > 0){
                        f[i][j] = f[i-1][j] + f[i][j-1] ;
                    }
                }
            }
        }
        return f[m-1][n-1] ;
    }
}