leetcode: Unique Paths II

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]
]

AC，还是得动态规划，dp[i][j] = dp[i-1][j] + dp[i][j-1]，注意加前先判断对应ob是不是为1，为1的就不加

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
if( obstacleGrid[0][0] || obstacleGrid[obstacleGrid.size()-1][obstacleGrid[0].size()-1])//如果左上角和右下角为1，那就肯定到不了了
return 0;
vector< vector< int > > dp( obstacleGrid.size(), vector< int>( obstacleGrid[0].size(), 0));
int i = 0;
for( ; i < obstacleGrid.size(); ++i){//ob[i][0]里为1，初始化时候，dp对应位子后面就都是0，因为只有一条路，有一个障碍就后面都到不了
if( obstacleGrid[i][0])
break;
}
for( int j = 0; j < i; ++j)
dp[j][0] = 1;
i = 0;
for( ; i < obstacleGrid[0].size(); ++i){//同理
if( obstacleGrid[0][i])
break;
}
for( int j = 0; j < i; ++j)
dp[0][j] = 1;
for( i = 1; i < obstacleGrid.size(); ++i){
for( int j = 1; j < obstacleGrid[0].size(); ++j){
if( obstacleGrid[i-1][j] && obstacleGrid[i][j-1])
dp[i][j] = 0;
else if( obstacleGrid[i-1][j])
dp[i][j] = dp[i][j-1];
else if( obstacleGrid[i][j-1])
dp[i][j] = dp[i-1][j];
else
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
return dp[obstacleGrid.size()-1][obstacleGrid[0].size()-1];
}
};

TLE

递归太慢

class Solution {
private:
int cnt = 0;
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
core( obstacleGrid, 0, 0, obstacleGrid.size(), obstacleGrid[0].size());
return cnt;
}
void core( vector< vector< int> > & ob, int row, int col, int m, int n){
if( row == m && col == n){
++cnt;
return;
}
else if( row >= m || col >= n){
return;
}
if( ob[row][col] == 1)
return;
core( ob, row+1, col, m, n);
core( ob, row, col+1, m, n);
return;
}
};