LeetCode 63. Unique Path II

最新推荐文章于 2024-03-21 18:04:53 发布

无名_1989

最新推荐文章于 2024-03-21 18:04:53 发布

阅读量139

点赞数

本文介绍了一个基于动态规划的算法，用于解决在存在障碍物的情况下计算从起点到终点的不同路径数量的问题。该算法通过修改原始网格来记录到达每个单元格的不同路径数，并考虑了障碍物的影响。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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) 
    {
        if(obstacleGrid[0][0] == 1 || 
                obstacleGrid[obstacleGrid.length-1][obstacleGrid[0].length-1] == 1) // obstacle at start point or finish point
            return 0;

        obstacleGrid[0][0] = -1; // start point
        
        for(int i=0; i<obstacleGrid.length; i++) // row
        {
            for(int j=0; j<obstacleGrid[0].length; j++) // column
            {
                // if this is not obstacle
                if(obstacleGrid[i][j] !=1)
                {
                    // get left: left is not obstacle 
                    if(j-1 >=0 && obstacleGrid[i][j-1] !=1)
                        obstacleGrid[i][j] += obstacleGrid[i][j-1];
                    // get top: top is not obstacle
                    if(i-1 >=0 && obstacleGrid[i-1][j] !=1)
                        obstacleGrid[i][j] += obstacleGrid[i-1][j];
                }
                
            }
        }
        
        return obstacleGrid[obstacleGrid.length-1][obstacleGrid[0].length-1] * -1;
    }
}