leetcode 063 —— Unique Paths II

最新推荐文章于 2020-07-09 17:26:02 发布

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本文探讨了在一个存在障碍物的网格中寻找从左上角到右下角的唯一路径数量的问题。采用动态规划的方法，详细阐述了算法的实现过程，并提供了一个C++示例代码。

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.

思路：依旧是动态规划

class Solution {
public:
	int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
		int m = obstacleGrid.size();
		int n = obstacleGrid[0].size();
		vector<vector<int>> num(m, vector<int>(n, 0));
		num[0][0] = 1 - obstacleGrid[0][0];
		if (num[0][0]==0) return 0; //表示第一个点就没有通过

		for (int i = 1; i < m; i++){
			if (obstacleGrid[i][0] == 1) num[i][0] = 0;
			else
				num[i][0] = num[i - 1][0];
		}
		for (int i = 1; i < n; i++){
			if (obstacleGrid[0][i] == 1) num[0][i] = 0;
			else
				num[0][i] = num[0][i-1];
		}

		for (int i = 1; i < m; i++){
			for (int j = 1; j < n; j++){
				if (obstacleGrid[i][j] == 1)
					num[i][j] = 0;
				else
					num[i][j] = num[i - 1][j] + num[i][j - 1];
			}
		}
		return num[m - 1][n - 1];
	}
};