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[].size();
int dp[][];
dp[][]=obstacleGrid[][]==?-:;
for(int i=;i<m;i++)
{
if(obstacleGrid[i][]==)
{
dp[i][]=-;
continue;
}
if(dp[i-][]==-) dp[i][]=-;
else dp[i][]=;
}
for(int j=;j<n;j++)
{
if(obstacleGrid[][j]==)
{
dp[][j]=-;
continue;
}
if(dp[][j-]==-) dp[][j]=-;
else dp[][j]=;
}
for(int i=;i<m;i++)
{
for(int j=;j<n;j++)
{
if(obstacleGrid[i][j]==)
{
dp[i][j]=-;
continue;
}
if(dp[i-][j]==-&&dp[i][j-]==-) dp[i][j]=-;
else if(dp[i-][j]==-&&dp[i][j-]!=-) dp[i][j]=dp[i][j-];
else if(dp[i-][j]!=-&&dp[i][j-]==-) dp[i][j]=dp[i-][j];
else if(dp[i-][j]!=-&&dp[i][j-]!=-) dp[i][j]=dp[i-][j]+dp[i][j-];
}
}
return dp[m-][n-]==-?:dp[m-][n-];
}
};
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[].size();
int dp[][];
//vector<vector<int>> dp(m,vector<int>(n));
dp[][]=obstacleGrid[][]==?:;
for(int i=;i<m;i++)
{
dp[i][]=obstacleGrid[i][]==?:dp[i-][];
}
for(int j=;j<n;j++)
{
dp[][j]=obstacleGrid[][j]==?:dp[][j-];
}
for(int i=;i<m;i++)
{
for(int j=;j<n;j++)
{
dp[i][j]=obstacleGrid[i][j]==?:dp[i-][j]+dp[i][j-];
}
}
return dp[m-][n-];
}
};