题目
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.
解题之法
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)); //注意初始化方法
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();
}
};
分析
这道题是之前那道 Unique Paths 不同的路径 的延伸，在路径中加了一些障碍物，还是用动态规划Dynamic Programming来解，不同的是当遇到为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。
