[leetcode] Unique Paths II

From : https://leetcode.com/problems/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

.

Note: m and n will be at most 100.

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

优化：

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

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if(obstacleGrid == null || obstacleGrid.length == 0) {
return 0;
}
int m = obstacleGrid.length, n = obstacleGrid[0].length;
int[] v = new int
;
for(int i=0; i<n && obstacleGrid[0][i]==0; ++i) {
v[i] = 1;
}

int flag = 1-obstacleGrid[0][0];
for(int i=1; i<m; ++i) {
if(flag==0 || obstacleGrid[i][0]==1) {
flag = 0;
}
v[0] = flag;
for(int j=1; j<n; ++j) {
v[j] = (1-obstacleGrid[i][j])*(v[j]+v[j-1]);
}
}

return v[n-1];
}
}