Leetcode - Unique Paths II

最新推荐文章于 2021-11-14 23:25:40 发布

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本文探讨了在网格中寻找不经过障碍物的独特路径的问题，详细介绍了使用动态规划算法解决此类问题的方法，包括如何初始化动态规划矩阵、如何更新矩阵元素以及最终计算网格终点的独特路径数量。

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.

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null)
            return 0;
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if (m == 0 || n == 0 || obstacleGrid[0][0] == 1)
            return 0;
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1)
                    dp[i][j] = 0;
                else if (i > 0 && j > 0) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                } else if (i > 0) {
                    dp[i][j] = dp[i - 1][j];
                } else if (j > 0) {
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
}