【leetcode】Unique Paths II

Unique Paths II

 Total Accepted: 22828 Total Submissions: 81414My Submissions

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.

把存在Obstacle的位置标记为-1，表示无法通行

 

 

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
        
        int dp[101][101];
        
        dp[0][0]=obstacleGrid[0][0]==1?-1:1;
        
        for(int i=1;i<m;i++)
        {
            if(obstacleGrid[i][0]==1) 
            {
                dp[i][0]=-1;
                continue;
            }
            if(dp[i-1][0]==-1) dp[i][0]=-1;
            else dp[i][0]=1;
        }
        
        for(int j=1;j<n;j++)
        {
            if(obstacleGrid[0][j]==1) 
            {
                dp[0][j]=-1;
                continue;
            }
            
            if(dp[0][j-1]==-1) dp[0][j]=-1;
            else dp[0][j]=1;
        }
        
        for(int i=1;i<m;i++)
        {
            for(int j=1;j<n;j++)
            {
                if(obstacleGrid[i][j]==1) 
                {
                    dp[i][j]=-1;
                    continue;
                }
                if(dp[i-1][j]==-1&&dp[i][j-1]==-1) dp[i][j]=-1;
                else if(dp[i-1][j]==-1&&dp[i][j-1]!=-1) dp[i][j]=dp[i][j-1];
                else if(dp[i-1][j]!=-1&&dp[i][j-1]==-1) dp[i][j]=dp[i-1][j];
                else if(dp[i-1][j]!=-1&&dp[i][j-1]!=-1) dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        
        return dp[m-1][n-1]==-1?0:dp[m-1][n-1];
    }
};

 

 

 

实际上不用标记也可以，dp[i][j]=0就表示了没有路

 

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
        
        int dp[101][101];
        //vector<vector<int>> dp(m,vector<int>(n));
        
        dp[0][0]=obstacleGrid[0][0]==1?0:1;
        for(int i=1;i<m;i++)
        {
            dp[i][0]=obstacleGrid[i][0]==1?0:dp[i-1][0];
        }
        for(int j=1;j<n;j++)
        {
            dp[0][j]=obstacleGrid[0][j]==1?0:dp[0][j-1];
        }
        for(int i=1;i<m;i++)
        {
            for(int j=1;j<n;j++)
            {
                dp[i][j]=obstacleGrid[i][j]==1?0:dp[i-1][j]+dp[i][j-1];
            }
        }
        
        return dp[m-1][n-1];
    }
};