leetcode63 Unique Paths II

最新推荐文章于 2018-07-05 19:20:46 发布

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最新推荐文章于 2018-07-05 19:20:46 发布

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原文链接：http://www.cnblogs.com/jsir2016bky/p/5105885.html

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左上角到右下角的不同路径数量的问题。该算法考虑了障碍物的存在，并通过更新路径计数来避免碰撞。

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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.

 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
 4         int ans=0;
 5         int m=obstacleGrid.size();
 6         if(!m)
 7             return ans;
 8         int n=obstacleGrid[0].size();
 9         
10         vector<int> jlist(n,0);
11         for(int i=n-1;i>=0;i--)
12         {
13             if(obstacleGrid[m-1][i]==1)
14                 break;
15             else
16                 jlist[i]=1;
17         }
18         
19         for(int i=m-2;i>=0;i--)
20         {
21             for(int j=n-1;j>=0;j--)
22             {
23                 if(j==n-1)
24                 {
25                     if(obstacleGrid[i][j]==1)
26                         jlist[j]=0;
27                 }
28                 else
29                 {
30                 
31                 if(obstacleGrid[i][j]==1)
32                     jlist[j]=0;
33                 else
34                     jlist[j]+=jlist[j+1];
35                     
36                 }
37                 
38                 
39             }
40         }
41         return jlist[0];
42     }
43 };