LeetCode:63. Unique Paths II 机器人所有可行路径_the robot starts in the upper left corner and has -优快云博客

本文链接：https://blog.youkuaiyun.com/qq_16234613/article/details/93369691

博客围绕机器人在带障碍物的 m x n 网格中寻找独特路径的问题展开。机器人只能向下或向右移动，障碍物和空白区域分别用 1 和 0 表示，还给出了示例输入和输出，探讨如何计算到达右下角的独特路径数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

试题
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

Right -> Right -> Down -> Down
Down -> Down -> Right -> Right

代码

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] count = new int[m][n];
        if(obstacleGrid[0][0]==0)
            count[0][0] = 1;
        else
            count[0][0] = 0;
        
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(i==0 && j==0) continue;
                if (obstacleGrid[i][j]==1){
                    count[i][j] = 0;
                    continue;
                }
                if(i==0 && j>0){
                    count[i][j] = count[i][j-1];
                    continue;
                }
                if(j==0 && i>0){
                    count[i][j] = count[i-1][j];
                    continue;
                }
                count[i][j] = count[i-1][j] + count[i][j-1];
            }
        }
        return count[m-1][n-1];
    }
}