(M)Dynamic Programming:63. Unique Paths II

这个题和Unique Paths差不多，就是在算dp[i][j]的时候，要考虑它的上边和左边是否有障碍物，只加没有障碍物的那个。最后要注意，如果右下角也就是目的地的位置是障碍物，那么要返回0.

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

看了大神的写法：当遇到为1的点，将该位置的dp数组中的值清零，其余和之前那道题并没有什么区别
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1)
return 0;
vector<vector<int> > dp(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(), 0));
for (int i = 0; i < obstacleGrid.size(); ++i) {
for (int j = 0; j < obstacleGrid[i].size(); ++j) {
if (obstacleGrid[i][j] == 1) dp[i][j] = 0;
else if (i == 0 && j == 0) dp[i][j] = 1;
else if (i == 0 && j > 0) dp[i][j] = dp[i][j - 1];
else if (i > 0 && j == 0) dp[i][j] = dp[i - 1][j];
else dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp.back().back();
}
};