63. Unique Paths II

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最新推荐文章于 2024-07-25 09:04:50 发布

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分类专栏： leetcode

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本文探讨了一个经典动态规划问题——在存在障碍的情况下寻找从起点到终点的不同路径数量。在一个由0和1组成的矩阵中，0代表可以通行的空间，而1则表示障碍物。通过动态规划的方法，我们能够有效地计算出所有可能的不重复路径总数。

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Follow up for “Unique Paths”:

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.

For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.

Note: m and n will be at most 100.
题目大意：在中间加了个障碍物，考虑的情况多了，但是还是动态规划

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int F[100][100];
        int rows = obstacleGrid.size(), cols = obstacleGrid[0].size();
        if(obstacleGrid[0][0] == 1)
            F[0][0] = 0;
        else 
            F[0][0] = 1;
        for(int j = 1; j < cols; ++j){
            if(obstacleGrid[0][j] == 1)
                F[0][j] = 0;
            else 
                F[0][j] = F[0][j - 1];
        }
        for(int i = 1; i < rows; ++i){
            if(obstacleGrid[i][0] == 1)
                F[i][0] = 0;
            else 
                F[i][0] = F[i - 1][0];
            for(int j = 1; j < cols; ++j){
                if(obstacleGrid[i][j] == 1)
                    F[i][j] = 0;
                else {
                    F[i][j] = F[i][j - 1] + F[i - 1][j];
                }
            }
        }

        return F[rows - 1][cols - 1];
    }
};