63. Unique Paths II


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];
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/63

类似题目:

Unique Paths

Unique Paths III

Minimum Path Cost in a Grid

Paths in Matrix Whose Sum Is Divisible by K

参考资料:

https://leetcode.com/problems/unique-paths-ii/

https://leetcode.com/problems/unique-paths-ii/discuss/23250/Short-JAVA-solution

https://leetcode.com/problems/unique-paths-ii/discuss/23248/My-C%2B%2B-Dp-solution-very-simple!

LeetCode All in One 题目讲解汇总(持续更新中…)

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