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 <  || m < ) return ;
         if (n ==  || m == ) return ;

         int[][] paths = new int[m][n];

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

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 ==  || obstacleGrid[].length == ) return ;

         int m = obstacleGrid.length;
         int n = obstacleGrid[].length;

         int[][] paths = new int[m][n];

         for (int i = ; i < m; i++) {
             for (int j = ; j < n; j++) {
                 if (obstacleGrid[i][j] == ) {
                     paths[i][j] = ;
                 } else if (i ==  && j == ) {
                     paths[i][j] = ;
                 } else if (i == ) {
                     paths[i][j] = paths[i][j - ];
                 } else if (j == ) {
                     paths[i][j] = paths[i - ][j];
                 } else {
                     paths[i][j] = paths[i - ][j] + paths[i][j - ];
                 }
             }
         }
         return paths[m - ][n - ];
     }
 }