Unique Paths I && II

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

DP：用一个二维数组，map[i][j] 到此点可能的路线数量，map[i][j] = map[i - 1][j] + map[i][ j - 1]。我们可以用一个一维数组来代替。

public class Solution {
    //Time: O(m*n)  Space: O(n)
    public int uniquePaths(int m, int n) {
        if (m <= 0 || n <= 0) {
            return 0;
        }
        
        int[] map = new int[n];
        Arrays.fill(map, 1);
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                map[j] = map[j] + map[j - 1];
            }
        }
        return map[n - 1];
    }
}

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.

基本和上一题一样，只需判断是否为障碍物，如是设为0。

public class Solution {
    //Time: O(m*n)  Space: O(n)
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1) {
            return 0;
        }
        
        int[] map = new int[obstacleGrid[0].length];
        map[0] = 1;
        for (int i = 0; i < obstacleGrid.length; i++) {
            for (int j = 0; j < obstacleGrid[0].length; j++) {
                if (obstacleGrid[i][j] == 1) {
                    map[j] = 0;
                } else if (j > 0) {
                    map[j] += map[j - 1];
                }
            }
        }
        return map[obstacleGrid[0].length - 1];
    }
}