Unique Paths(62,63)

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?

思路：

这是一个dp问题。

从最开始，rebort只能向下或者向右。我们假设到达point（i，j）的路径数被设置为p(i,j),那么状态方程：P[i][j] = P[i - 1][j] + P[i][j - 1].

The boundary conditions of the above equation occur at the leftmost column (P[i][j - 1] does not exist) and the uppermost row (P[i - 1][j] does not exist). These conditions can be handled by initialization (pre-processing) --- initialize P[0][j] = 1, P[i][0] = 1 for all valid i, j. Note the initial value is 1 instead of 0!

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

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.

Note: m and n will be at most 100.

总的思路和第一道类似，也是一个dp问题，公式还是：P[i][j] = P[i - 1][j] + P[i][j - 1].现在有一个问题就是，如果（i，j）这个点是障碍物，那么到达（i，j）的路径要设置为0.因为不可达。如果不为0，那么：P[i][j] = P[i - 1][j] + P[i][j - 1].

剩下的就是初始条件的设定。如果（0，0）为障碍物，那么直接p(0,0)为0不可达。

然后利用P[i][j] = P[i - 1][j] + P[i][j - 1]. 初始化p(m,0)和p（0,n）

public class Solution {

    // public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    //         int m = obstacleGrid.length;
    //         int n = obstacleGrid[0].length;
    
    //         obstacleGrid[0][0]^=1;
    //         for(int i = 1;i<m;i++){
    //             obstacleGrid[i][0]=(obstacleGrid[i][0]==1)? 0:obstacleGrid[i-1][0];
    //         }
            
    //         for(int j = 1;j<n;j++){
    //             obstacleGrid[0][j] =(obstacleGrid[0][j]==1)? 0: obstacleGrid[0][j-1];
    //         }
    //         for(int i = 1;i<m;i++){
    //             for(int j =1;j<n;j++){
    //                 obstacleGrid[i][j] =(obstacleGrid[i][j]==1)? 0: obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
    //             }
    //         }
    //         return obstacleGrid[m-1][n-1];
    //     }
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] map= new int[m][n];
        map[0][0]=obstacleGrid[0][0]^1;
        for(int i = 1;i<m;i++){
                map[i][0]=(obstacleGrid[i][0]==1)? 0:map[i-1][0];
        }
        for(int j = 1;j<n;j++){
            map[0][j] =(obstacleGrid[0][j]==1)? 0: map[0][j-1];
        }
        for(int i = 1;i<m;i++){
            for(int j =1;j<n;j++){
                map[i][j] =(obstacleGrid[i][j]==1)? 0: map[i-1][j]+map[i][j-1];
            }
        }
        return map[m-1][n-1];
        
    
    }    
}

总结：dp问题关键是找状态方程。 设置初始条件。剩下的按照题目一次添加额外条件。