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  • 【leetcode刷题笔记】Unique Paths II

    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.


    题解:和http://www.cnblogs.com/sunshineatnoon/p/3798167.html非常相似,要注意两点:

    1. 有障碍的地方能走到的方法数目是0;
    2. 初始化结果矩阵的时候,如果第一行(列)的某个位置有障碍,那么这一行(列)该元素后面所有的元素都是0,不能置为1;比如给定障碍矩阵[1,0,0],那么初始化以后的矩阵应该为[0,0,0],而不是[0,1,1],因为(0,0)处有障碍,那么该矩阵的任何位置都是无法到达的。

    代码如下:

     1 public class Solution {
     2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
     3         int m = obstacleGrid.length;
     4         int n = obstacleGrid[0].length;
     5         if(m==0 && n == 0)
     6             return 0;
     7         
     8         int[][] PathNum = new int[m][n];
     9         for(int i = 0;i < m;i++)
    10             if(obstacleGrid[i][0] != 1)
    11                 PathNum[i][0] = 1;
    12             else 
    13                 break;
    14         for(int i = 0;i < n;i++)
    15             if(obstacleGrid[0][i] != 1)
    16                 PathNum[0][i] = 1;
    17             else
    18                 break;
    19         
    20         for(int i = 1;i < m;i++)
    21             for(int j = 1;j < n;j++){
    22                 if(obstacleGrid[i][j] != 1)
    23                     PathNum[i][j] = PathNum[i-1][j]+PathNum[i][j-1]; 
    24             }
    25         
    26         return PathNum[m-1][n-1];
    27     }
    28 }
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  • 原文地址:https://www.cnblogs.com/sunshineatnoon/p/3858029.html
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