leetcode 63. Unique Paths II

最新推荐文章于 2024-07-26 10:47:39 发布

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最新推荐文章于 2024-07-26 10:47:39 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种使用动态规划解决带有障碍物的网格中从起点到终点的唯一路径数量的方法。针对不同情况，如起点即为障碍等进行了特殊处理。

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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.

这题的测试好恶心，居然左上角还有1的情况，请问机器人怎么站

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid==null || obstacleGrid.length==0) return 0;
        int w = obstacleGrid.length, h = obstacleGrid[0].length;
        int[][] dp = new int[w][h];
        // dp[0][0] = 1;
        if(obstacleGrid[0][0] ==1) return 0;
        dp[0][0] = 1;
        for(int i=1; i<w; i++){
            if(obstacleGrid[i][0]==1){
                dp[i][0] = 0;
            }else{
                dp[i][0] = dp[i-1][0];
            }
        }

        for(int j=1; j<h; j++){
            if(obstacleGrid[0][j]==1){
                dp[0][j] = 0;
            }else{
                dp[0][j] = dp[0][j-1];
            }
        }

        for(int i=1; i<w; i++){
            for(int j = 1; j<h; j++){
                if(obstacleGrid[i][j]==1){
                    dp[i][j]=0;
                }else{
                    dp[i][j] = dp[i-1][j]+dp[i][j-1];
                }
            }
        }

        return dp[w-1][h-1];
    }
}