8.2 Robot in a Grid
3960 단어 dp
O(mn) space code:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
// write your code here
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int> > dp(m+1,vector<int>(n+1,0));
dp[0][1] = 1;
for(int i = 1; i <= m; ++i){
for(int j = 1; j <= n; ++j){
if(!obstacleGrid[i-1][j-1]) dp[i][j] = dp[i-1][j]+dp[i][j-1];
}
}
return dp[m][n];
}
O(n) space;
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<int> dp(n,0);
for(int j = 0; j<n; ++j){
if(obstacleGrid[0][j] != 1) dp[j] = 1;
else break;// if there is a 1: which means the grid after it cannot be reach at initialization time.
}
for(int i = 1; i<m; ++i){
dp[0] = (obstacleGrid[i][0] == 1) ? 0 : dp[0];
for(int j = 1; j<n; ++j){
dp[j] = (obstacleGrid[i][j] == 1) ? 0 : dp[j] + dp[j-1];
}
}
return dp[n-1];
}