首页 > 代码库 > 【LeetCode】063. Unique Paths II
【LeetCode】063. Unique Paths II
题目:
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:
Solution 1 ()(未优化)
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { if(obstacleGrid.empty() || obstacleGrid[0].empty()) return 0; int m = obstacleGrid.size(), n = obstacleGrid[0].size(); vector<int> dp(n,0); dp[0] = 1; for(int i=0; i<m; ++i) { for(int j=0; j<n; ++j) { if(obstacleGrid[i][j] == 1) dp[j] = dp[j-1]; else dp[j] += dp[j-1]; } } return dp[n-1]; } };
Solution 2 ()
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { if(obstacleGrid.empty() || obstacleGrid[0].empty()) return 0; int m = obstacleGrid.size(), n = obstacleGrid[0].size(); vector<int> dp(n,0); dp[0] = 1; for(int i=0; i<m; ++i) { for(int j=0; j<n; ++j) { if(obstacleGrid[i][j] == 1) dp[j] = 0; else { if(j > 0) dp[j] += dp[j-1]; } } } return dp[n-1]; } };
【LeetCode】063. Unique Paths II
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。