LeetCode Unique Paths II

障碍路径计数

最新推荐文章于 2019-05-17 12:08:12 发布

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本文介绍了一个Unique Paths问题的进阶版，即在网格中存在障碍物的情况下，计算从起点到终点的不同路径数量。文章提供了C++实现代码，通过动态规划的方法解决了这一问题。

题目：

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) {
		int m = obstacleGrid.size();
		int n = obstacleGrid[0].size();
		f[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;
		for (int i = 1; i < m; i++)
			f[i][0] = obstacleGrid[i][0] == 0 ? f[i-1][0] : 0;
		for (int j = 1; j < n; j++)
			f[0][j] = obstacleGrid[0][j] == 0 ? f[0][j-1] : 0;
		for (int i = 1; i < m; i++) {
			for (int j = 1; j < n; j++) {
				f[i][j] = obstacleGrid[i][j] == 0 ? f[i - 1][j] + f[i][j - 1] : 0;
			}
		}
		return f[m - 1][n - 1];
	}
private:
	int f[100][100];
};