leetcode 63. Unique Paths II

最新推荐文章于 2024-07-26 10:47:39 发布

moli2333

最新推荐文章于 2024-07-26 10:47:39 发布

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本文介绍了一个基于动态规划的方法来计算在一个包含障碍物的网格中从起点到终点的不同路径数量。当网格中的某些位置被标记为障碍物时，该算法能够有效地找出所有可能的绕过这些障碍到达目的地的路径。

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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.

上题：
这里写图片描述

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        if(m == 0) return 0;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        int i,j;
        for(int index = m*n-1;index >= 0;index--){
            i = index / n;
            j = index % n;
            if(obstacleGrid[i][j] == 1){ 
                dp[i][j] = 0;
            }
            else if(index == m*n-1) dp[i][j] = 1;
            else{
                dp[i][j] = 0;
                if(i+1 < m) dp[i][j] += dp[i+1][j];
                if(j+1 < n) dp[i][j] += dp[i][j+1];
            }

        }
        return dp[0][0];
    }

加入一个逻辑即可：

if(obstacleGrid[i][j] == 1){ 
                dp[i][j] = 0;
            }