[刷题]Unique Paths II

最新推荐文章于 2024-04-01 00:43:20 发布

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本文介绍了解决带有障碍物的独特路径问题的算法，通过使用动态规划方法，针对二维网格计算从起点到终点的唯一路径数量。

[LintCode]Unique Paths II

public class Solution {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // 2015-05-14
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){
            return 0;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] sum = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    sum[i][j] = 0;
                    continue;
                }
                if (i == 0 && j == 0) {
                    sum[i][j] = 1;
                } else if (i == 0 && j != 0) {
                    sum[i][j] = sum[i][j - 1]; 
                } else if (i != 0 && j == 0) {
                    sum[i][j] = sum[i - 1][j];
                } else {
                    sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
                }
            }
        }
        return sum[m - 1][n - 1];
    }
}