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]
]

The total number of unique paths is

2

.

Note: m and n will be at most 100.

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class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {

vector<vector<int>> dp(obstacleGrid.size(), vector<int>(obstacleGrid[0].size()));

dp[0][0] = obstacleGrid[0][0] == 1? 0:1;

for (int i=1; i<obstacleGrid.size(); i++)
{
if (obstacleGrid[i][0] == 1)
{
dp[i][0] = 0;
}
else
{
dp[i][0] = dp[i-1][0];
}
}

for (int j=1; j<obstacleGrid[0].size(); j++)
{
if (obstacleGrid[0][j] == 1)
{
dp[0][j] = 0;
}
else
{
dp[0][j] = dp[0][j-1];
}
}

for (int i=1; i<obstacleGrid.size(); i++)
{
for (int j=1; j<obstacleGrid[0].size(); j++)
{
if (obstacleGrid[i][j] == 1)
{
dp[i][j] = 0;
}
else
{
dp[i][j] = dp[i-1][j]+dp[i][j-1];
}
}
}

return dp[obstacleGrid.size()-1][obstacleGrid[0].size()-1];
}
};

一个更有效率的解法：

int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int h = obstacleGrid.size();
if(h == 0) return 0;
int w = obstacleGrid[0].size();
if(w == 0) return 0;
if(obstacleGrid[0][0]) return 0;

// first cell has 1 path
obstacleGrid[0][0] = 1;

// first row all are '1' until obstacle (from left only)
for(int i=1; i<w; i++){
obstacleGrid[0][i] = obstacleGrid[0][i] ? 0 : obstacleGrid[0][i-1];
}

for(int j=1; j<h; j++){
// first column is like first row (from top only)
obstacleGrid[j][0] = obstacleGrid[j][0] ? 0 : obstacleGrid[j-1][0];

// others are up+left
for(int i=1; i<w; i++){
obstacleGrid[j][i] = obstacleGrid[j][i] ? 0 : obstacleGrid[j-1][i] + obstacleGrid[j][i-1];
}
}

return obstacleGrid[h-1][w-1];
}