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63. Unique Paths II(有障碍的路径 动态规划)

时间:2018-03-09 18:38:51      阅读:241      评论:0      收藏:0      [点我收藏+]


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.

 

 

 1 class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         int rows = obstacleGrid.length;
 4         int cols = obstacleGrid[0].length;
 5         for(int i = 0; i < rows;i++){
 6             for (int j = 0; j < cols ;j++){
 7                 if(obstacleGrid[i][j]==1)
 8                     obstacleGrid[i][j] = 0;
 9                 else if (i==0&& j==0) 
10                     obstacleGrid[i][j] = 1;
11                 else if (i==0) 
12                     obstacleGrid[i][j] = obstacleGrid[i][j-1]*1; //边界,没有路径了,要么是0,要么是1
13                 
14                 else if (j == 0)
15                     obstacleGrid[i][j] = obstacleGrid[i-1][j]*1;
16                 else 
17                     obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
18             }
19         }
20         return obstacleGrid[rows-1][cols-1];
21     }
22 }

 

63. Unique Paths II(有障碍的路径 动态规划)

原文:https://www.cnblogs.com/zle1992/p/8535285.html

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