LeetCode #63 Unique Paths II 不同路径 II

63 Unique Paths II 不同路径 II

Description:
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?

Paths II

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:

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:

Right -> Right -> Down -> Down
Down -> Down -> Right -> Right

题目描述:
一个机器人位于一个 m x n 网格的左上角（起始点在下图中标记为“Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

路径 II

网格中的障碍物和空位置分别用 1 和 0 来表示。

说明：
m 和 n 的值均不超过 100。

示例 :

示例 1:

输入:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：

向右 -> 向右 -> 向下 -> 向下
向下 -> 向下 -> 向右 -> 向右

思路:

动态规划

边界处理, 如果入口就有障碍, 直接返回
第一个元素设置为 1, 表示可达, 初始化第一行/列, 如果可达, 标记为 1否则标记为 0
对除第一行/列进行遍历, 如果该元素没有障碍,则 dp[i][j] = dp[i - 1][j] + dp[i][j - 1], 否则标记为 0表示不可达
返回最后一个元素, dp数组可以直接用 obstacleGrid数组代替
时间复杂度O(mn), 空间复杂度O(1)

代码:
C++:

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

Java:

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid.length == 0 || obstacleGrid[0].length == 0 || obstacleGrid[0][0] == 1) return 0;
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        obstacleGrid[0][0] = 1;
        for (int i = 1; i < m; i++) obstacleGrid[i][0] = (obstacleGrid[i - 1][0] == 1 && obstacleGrid[i][0] == 0) ? 1 : 0;
        for (int i = 1; i < n; i++) obstacleGrid[0][i] = (obstacleGrid[0][i - 1] == 1 && obstacleGrid[0][i] == 0) ? 1 : 0;
        for (int i = 1; i < m; i++) for (int j = 1; j < n; j++) obstacleGrid[i][j] = obstacleGrid[i][j] == 0 ? obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1] : 0;
        return obstacleGrid[m - 1][n - 1];
    }
}

Python:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        if not obstacleGrid or not obstacleGrid[0] or obstacleGrid[0][0] == 1:
            return 0
        m, n, obstacleGrid[0][0] = len(obstacleGrid), len(obstacleGrid[0]), 1
        for i in range(1, m):
            obstacleGrid[i][0] = 1 if obstacleGrid[i][0] == 0 and obstacleGrid[i - 1][0] == 1 else 0
        for i in range(1, n):
            obstacleGrid[0][i] = 1 if obstacleGrid[0][i] == 0 and obstacleGrid[0][i - 1] == 1 else 0
        for i in range(1, m):
            for j in range(1, n):
                obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1] if not obstacleGrid[i][j] else 0
        return obstacleGrid[-1][-1]

最后编辑于：2021.04.03 09:00:50