开始刷leetcode day60: Unique Paths II

最新推荐文章于 2025-08-26 21:13:57 发布

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本文探讨了在网格中加入障碍物后，独特的路径数量如何计算。通过实例展示了解决此类问题的方法。

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.

Java:

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if(obstacleGrid == null) return 0;
int row = obstacleGrid.length;
int col = obstacleGrid[0].length;


int[][] destination = new int[row][col];
if(obstacleGrid[0][0]==1) return 0;
destination[0][0] = 1;
for(int i=1; i<row; i++)
{
if(destination[i-1][0] != 0 && obstacleGrid[i][0] != 1) destination[i][0] = 1;
else destination[i][0] = 0;
}

for(int j=1; j<col; j++)
{
if(destination[0][j-1]!=0 && obstacleGrid[0][j] != 1) destination[0][j] = 1;
else destination[0][j] = 0;
}

for(int i=1; i<row; i++)
{
for(int j=1; j<col; j++)
{
if(obstacleGrid[i][j] == 1) destination[i][j] = 0;
else destination[i][j] = destination[i-1][j] + destination[i][j-1];
}
}


return destination[row-1][col-1];
}
}