[Leetcode] Unique Paths II

最新推荐文章于 2019-05-17 12:08:12 发布

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最新推荐文章于 2019-05-17 12:08:12 发布

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分类专栏： LeetCode Algorithm

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本文介绍了一个基于动态规划的算法，用于解决在存在障碍物的情况下计算从起点到终点的不同路径数量的问题。该算法考虑了障碍物对路径的影响，并通过逐行更新路径计数来高效地求解。

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Unique Paths II Mar 29 '12 4573 / 11497

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.

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