Unique Paths II

Desctiption

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.

解题思路：这道题是在前面unique path的基础上进行了修改，解题用的方法也是动态规划，假设f（i，j）表示从起点到（i，j）这个点的不同路径数，前提是（i，j）是可达的，即（i，j）这个点不不是障碍物，由于从上一个点到达（i，j）这个点只有两种可能性，要么是从（i-1，j）这个点往下走，要么是从（i，j-1）这个点往右走，所以f（i，j）=f（i-1，j）+f（i，j-1），但是还要考虑（i-1，j）和（i，j-1）这两个点是不是可到达的。不妨这样认为，到一个为障碍物的点的路径数为0，那么就对计算下一个要到达的点的路径数没有影响了。代码如下：

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
vector<vector<int>> res(obstacleGrid.size(), vector<int>(obstacleGrid[0].size()));
bool flag = true;
for (int i = 0; i < obstacleGrid[0].size(); i++) {
if (!flag) { //初始化第一行的点的路径数，如果第一行出现了障碍物，那么障碍物后面的点都不可达，因为只能往右和往下走
res[0][i] = 0;
continue;
}
if (obstacleGrid[0][i] == 0)
res[0][i] = 1;
else {
flag = false;
res[0][i] = 0;
}
}
flag = true;
for (int i = 0; i < obstacleGrid.size(); i++) {
if (!flag) {//初始化第一列的点的路径数，如果出现障碍物，障碍物后面的点同样不可达。
res[i][0] = 0;
continue;
}
if (obstacleGrid[i][0] == 0)
res[i][0] = 1;
else {
res[i][0] = 0;
flag = false;
}
}
for (int i = 1; i < obstacleGrid.size(); i++) {
for (int j = 1; j < obstacleGrid[0].size(); j++) {
if (obstacleGrid[i][j] == 1) {
res[i][j] = 0;
continue;
}
else {
res[i][j] = res[i - 1][j] + res[i][j - 1];
}
}
}
return res[obstacleGrid.size() - 1][obstacleGrid[0].size() - 1];
}
};