2022.01.12 - SX10-07.不同路径 II

1. 题目2. 思路

(1) 动态规划 3. 代码

若网格[i,j]处有障碍，则到达网格[i,j]的路径数为0，否则，到达网格[i,j]的路径数为f(i,j)=f(i-1,j)+f(i,j-1)。初始化时，对于最左边和最上面的网格，若到达前一个网格的路径数为0，则之后网格的路径数均为0。 3. 代码

public class Test {
    public static void main(String[] args) {
    }
}

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