算法第14周Unique Paths II[medium]

Description

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.

Solution

这个题与上一题的区别是多了障碍物，所以当位置（i,j）为1时，它的路径数为0；

只需在上一题基础上加以修改即可。

同时需注意边界。

当第0行或者第0列出现障碍物时，其右边或者下边所有都不能到达。

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size() , n = obstacleGrid[0].size();
int dp[m]
;
bool f = true;
for (int i = 0; i < n; i++) {
if (!f || obstacleGrid[0][i] == 1) {
f = false;
dp[0][i] = 0;
} else {
dp[0][i] = 1;
}

}
f = true;
for (int i = 0; i < m; i++) {
if (!f || obstacleGrid[i][0] == 1) {
f = false;
dp[i][0] = 0;
} else {
dp[i][0] = 1;
}
}
for (int i = 1; i < m; i++) {
for (int j =
8f08
1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
dp[i][j] = 0;
} else {
dp[i][j] = dp[i-1][j]+dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
};