leetcode -- 63. Unique Paths II

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最新推荐文章于 2024-10-01 18:52:56 发布

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博客围绕机器人在带障碍的 m x n 网格中寻找从左上角到右下角的唯一路径数量展开。通过动态规划解决该问题，与普通网格路径问题不同的是，有障碍处的动态规划值需设为 0，还提及参考了柳神的精简代码。

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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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/unique-paths-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路：

1. 跟62题差不多，但是增加了障碍，障碍出的dp得设置为为0。

2. 看似差不多，但其实坑还挺多的。

3. 写了很多逻辑不太通。看了下柳神的代码，精简！！

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int row = obstacleGrid.size(); //行
        int col = obstacleGrid[0].size(); //列
        long long dp[100][100] = {0};
        for(int i = 0; i < row; i++){     
            for(int j = 0; j < col; j++){
                if(obstacleGrid[i][j] == 1) dp[i][j] = 0;
                else if(i == 0 && j == 0) dp[i][j] = 1;
                else if(i == 0) dp[i][j] = dp[i][j-1];
                else if(j == 0) dp[i][j] = dp[i-1][j];
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
            
        }
        return dp[row-1][col-1];
        
    }
};