63. 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]
]

The total number of unique paths is 2.

1 time is O(mn) ,space is O(mn)

path is m*n

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

2 time is O(mn),space is O(mn),but code is more concise

path is (m+1)*(n+1)

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

3 code more concise,space is O(m)

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