LeetCode 62 Unique Paths II

最新推荐文章于 2025-08-24 15:22:07 发布

mlweixiao

最新推荐文章于 2025-08-24 15:22:07 发布

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分类专栏： leetcode 文章标签： java leetcode

本文链接：https://blog.youkuaiyun.com/mlweixiao/article/details/39001637

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本文介绍了一种计算网格中从左上角到右下角唯一路径数量的方法，当网格中存在障碍物时。文章提供了一个Java实现示例，通过动态规划的方式解决了这一问题。

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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.

思路：与LeetCode 61 Unique Paths 的思路一样点http://blog.youkuaiyun.com/mlweixiao/article/details/37569159

public class Solution {
	public int uniquePathsWithObstacles(int[][] obstacleGrid) {
		int m = obstacleGrid.length;
		int n = obstacleGrid[0].length;

		int[][] grid = new int[m][n];
		int i, j;
		boolean flag = false;

		for (i = 0; i < m; i++) {
			if (obstacleGrid[i][0] == 1)
				flag = true;
			if (!flag)
				grid[i][0] = 1;
		}

		flag = false;
		for (i = 0; i < n; i++) {
			if (obstacleGrid[0][i] == 1)
				flag = true;
			if (!flag)
				grid[0][i] = 1;
		}

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