another classic DP problem.
2D grid, m*n, each time can only move right or down, how many unique paths from left top to right bottom?
dp[i] [j] represents for the number of ways from 0,0 to i,j

当然 下面的code的空间复杂度可以优化到O(min(m,n))

class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) return 0;
        
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

63
now we mark 1 in grid as obstacles, and 0 as space, then now how many ways?

dp[i] [j] still represents for the number of ways here, but remember the obstacles, the core state transit functions are the same as before.
so easy peasy.

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