Leetcode 63. Unique Paths II

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/qinduanyinghua/p/5733271.html

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#数据结构与算法

本文介绍了一种求解在网格中存在障碍物时从起点到终点的不同路径数量的算法。利用动态规划的方法，通过初始化边界条件并递推计算每个格子可达路径的数量，最终得出总的不同路径数。

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.

思路：动态规划题。

一开始边界的初始化很重要，其他的位置，如果是1，则置为0，否则的dp[i][j] = dp[i-1][j] + dp[i][j-1];

 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
 4         if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
 5             return 0;
 6         int dp[101][101], i, j;
 7         if(obstacleGrid[0][0] == 0){
 8             dp[0][0] = 1;
 9         }
10         else{
11             return 0;
12         }
13         
14         for(i = 1; i < obstacleGrid.size(); i++){
15             if(obstacleGrid[i][0] == 0 && dp[i-1][0] == 1)
16                 dp[i][0] = 1;
17             else
18                 dp[i][0] = 0;
19         }
20         
21         for(j = 1; j < obstacleGrid[0].size(); j++){
22             if(obstacleGrid[0][j] == 0 && dp[0][j-1] == 1)
23                 dp[0][j] = 1;
24             else
25                 dp[0][j] = 0;
26         }
27         
28         for( i = 1; i < obstacleGrid.size(); i++){
29             for(j = 1; j < obstacleGrid[0].size(); j++){
30                 if(obstacleGrid[i][j] == 0)
31                     dp[i][j] = dp[i-1][j] + dp[i][j-1];
32                 else
33                     dp[i][j] = 0;
34             }
35         }
36         
37         return dp[i-1][j-1];
38         
39     }
40 };