[leetcode 63] Unique Paths II

障碍路径计数

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本文介绍了一个基于动态规划算法的问题——在包含障碍物的网格中计算从起点到终点的不同路径数量。文章提供了一段C++代码实现，展示了如何通过动态规划解决这一问题。

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) {
        const int m = obstacleGrid.size();
        if (m == 0) {
            return 0;
        }
        const int n = obstacleGrid[0].size();
        if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) {
            return 0;
        }
        int f[n+1];
        fill_n(&f[0], n+1, 1);
        f[0] = 0;
        int i = 0;
        for (i = 1; i <= n; i++) {
            if (obstacleGrid[0][i-1] == 1) {
                f[i] = 0;
                break;
            }
        }
        for (int j = i; j <= n; j++) {
            f[j] = 0;
        }
        for (int i = 2; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                f[j] = obstacleGrid[i-1][j-1]?0:f[j]+f[j-1];
            }
        }
        return f[n];
    }
};