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  • Unique Paths II

    Description:

    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.

    Code:

     1     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
     2         int m = obstacleGrid.size();
     3         int n = obstacleGrid[0].size();
     4 
     5     //path[i][j]表示(0,0)号元素到(i,j)号元素的路径数,所以最终结果为path[m-1][j-1]
     6 
     7         int path[MAX][MAX] = {0};
     8         path[0][0] = (obstacleGrid[0][0] == 1)?0:1;
     9 
    10         for (int i = 1; i < m; ++i)
    11         {
    12             path[i][0] = (obstacleGrid[i][0]==1)?0:path[i-1][0];
    13         }
    14 
    15         for (int i = 1; i < n; ++i)
    16         {
    17             path[0][i] = (obstacleGrid[0][i]==1)?0:path[0][i-1];
    18         }
    19         
    20         for (int i = 1; i < m; ++i)
    21         {
    22             for (int j = 1; j < n; ++j)
    23             {
    24                 path[i][j] = (obstacleGrid[i][j] == 1)?0:(path[i-1][j] + path[i][j-1]);
    25             }
    26         }
    27         return path[m-1][n-1];
    28     }    
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  • 原文地址:https://www.cnblogs.com/happygirl-zjj/p/4582137.html
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