leetcode--Unique Paths II

最新推荐文章于 2020-08-18 21:35:00 发布

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最新推荐文章于 2020-08-18 21:35:00 发布

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原文链接：http://www.cnblogs.com/averillzheng/p/3774551.html

本文探讨了在存在障碍物的网格中寻找唯一路径的问题，提供了一种动态规划算法的解决方案，通过构建二维数组记录从起点到各点的路径数，有效地解决了问题。

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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.

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

转载于:https://www.cnblogs.com/averillzheng/p/3774551.html