[LeetCode] 63. Unique Paths II 不同路径之二

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[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

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Constraints:

m == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
obstacleGrid[i][j] is 0 or 1.

这道题是之前那道 Unique Paths 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，使用一个二维的 dp 数组，大小为 (m+1) x (n+1)，这里的 dp[i][j] 表示到达 (i-1, j-1) 位置的不同路径的数量，那么i和j需要更新的范围就是 [1, m] 和 [1, n]。状态转移方程跟之前那道题是一样的，因为每个位置只能由其上面和左面的位置移动而来，所以也是由其上面和左边的 dp 值相加来更新当前的 dp 值，如下所示：

dp[i][j] = dp[i-1][j] + dp[i][j-1]

这里就能看出来初始化 d p数组的大小为 (m+1) x (n+1)，是为了 handle 边缘情况，当i或j为0时，减1可能会出错。当某个位置是障碍物时，其 dp 值为0，直接跳过该位置即可。这里还需要初始化 dp 数组的某个值，使得其能正常累加。当起点不是障碍物时，其 dp 值应该为1，即dp[1][1] = 1，由于其是由 dp[0][1] + dp[1][0] 更新而来，所以二者中任意一个初始化为1即可。由于之后 LeetCode 更新了这道题的 test case，使得使用 int 型的 dp 数组会有溢出的错误，所以改为使用 long 型的数组来避免 overflow，代码如下：

解法一：

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

或者我们也可以使用一维 dp 数组来解，省一些空间，参见代码如下：

解法二：

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