A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
題意：從左上角走到右下角，如果有障礙，總走法是多少。
思路：還是動態規劃，如果是障礙，dp 矩陣中該位置的走法設爲 0 即可。和洛谷的【過河卒】很像。
代碼：
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid.empty() || obstacleGrid[0].empty()) return 0;
const auto &v = obstacleGrid;
int m = v.size(), n = v[0].size();
vector<vector<long long>> dp(m, vector<long long>(n, 0));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (v[i][j] == 1) dp[i][j] = 0; //設置障礙
else {
if (!i && !j) dp[0][0] = 1; //從(0,0)開始
else if (!i) dp[i][j] = dp[i][j - 1]; //第0行
else if (!j) dp[i][j] = dp[i - 1][j]; //第0列
else dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};
效率：
執行用時：8 ms, 在所有 C++ 提交中擊敗了30.49% 的用戶
內存消耗：8 MB, 在所有 C++ 提交中擊敗了100.00% 的用戶