63. Unique Paths II

最新推荐文章于 2025-08-15 18:53:39 发布

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最新推荐文章于 2025-08-15 18:53:39 发布

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分类专栏： # 算法实践文章标签： DP c++

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本文介绍了一种解决网格中存在障碍物时寻找从左上角到右下角唯一路径数量的问题。通过动态规划方法，定义状态转移方程并实现算法。

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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.

题意：给定一个二维数组，其中1表示障碍点，0表示无障碍点，求左上角到右下角有多少路径。

思路：dp[m][n] = obstancleGrid[m][n]==1? 0 : dp[m-1][n] + dp[m][n-1];

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