Unique Paths | & ||

Unique Paths I

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?

分析：

用A[i][j]表示到达点i,j可能的走法。 对于点A[i][j]，它可以从上一个格子下来，也可以从左边那个格子过来。所以A[i][j] = A[i-1][j] + A[i][j-1].

public class Solution {
    /**
     * @param n, m: positive integer (1 <= n ,m <= 100)
     * @return an integer
     */
    public int uniquePaths(int m, int n) {
        if (n < 0 || m < 0) return 0;
        if (n == 1 || m == 1) return 1;
        
        int[][] paths = new int[m][n];

        for (int i = 0; i < paths.length; i++) {
            for (int j = 0; j < paths[0].length; j++) {
                if (i == 0 || j == 0) {
                    paths[i][j] = 1;
                } else {
                    paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
                }
            }
        }
        return paths[m - 1][n - 1];
    }
}

Another solution:

 (For clarity, we will solve this part assuming an X+1 by Y+1 grid)

Each path has X+Y steps. Imagine the following paths:

X X Y Y X (we move right on the first 2 steps, then down on the next 2, then right  for the last step)

X Y X Y X (we move right, then down, then right, then down, then right)

…

Each path can be fully represented by the moves at which we move right. That is, if I were to ask you which path you took, you could simply say “I moved right on step 3 and 4.” Since you must always move right X times, and you have X + Y total steps, you have to pick X times to move right out of X+Y choices. Thus, there are C(X, X+Y) paths (eg, X+Y choose X).

Unique Paths II

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.

 

Example

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.

分析：

原理同上，没有任何区别。

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