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LeetCode——Unique Paths II

2014年11月09日 ⁄ 综合 ⁄ 共 1040字 ⁄ 字号 评论关闭

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.

»
Solve this problem

在Unique Paths的基础上稍作修改即可,在grid上为1的点,其f值为0,表示无论如何都没法到达。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) {
            return 0;
        }
        
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        int f[m][n];
        memset(f, 0, sizeof(int) * m * n);
        f[0][0] = 1;
        
        for (int i = 1; i < m; i++) {
            f[i][0] = obstacleGrid[i][0] == 1 ? 0 : f[i - 1][0];
        }
        for (int i = 1; i < n; i++) {
            f[0][i] = obstacleGrid[0][i] == 1 ? 0 : f[0][i - 1];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    f[i][j] = 0;
                }
                else {
                    f[i][j] += obstacleGrid[i - 1][j] == 1 ? 0 : f[i - 1][j];
                    f[i][j] += obstacleGrid[i][j - 1] == 1 ? 0 : f[i][j - 1];                    
                }
            }
        }
        
        return f[m - 1][n - 1];
    }
};

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