LeetCode刷题笔录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.

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.

这题比Unique Paths多了一点限制条件，因此不是一个简单的组合数的问题了。

自然的思路是动态规划。A[m][n]是到达(m,n)点得Unique Path数目。则有：

A[m][n] = 0 if obstacleGrid[m][n] = 1;

A[m][n] = A[m - 1][n] + A[m][n - 1] if obstacleGrid[m][n] = 0.

意思是如果这里有障碍，那么根本没有一条path能走到这里。如果这里没有障碍，那么可以从这个格子的上面或者左边走过来。

Base case是第一行和第一列。对于A[i][0],只要某个obstacleGrid[i][0]=1，则后面的所有A[i][0]=0.对于A[0][j]同理。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] A = new int[m][n];
        
        boolean blocked = false;
        for(int i = 0; i < m; i++){
            if(obstacleGrid[i][0] == 1){
                blocked = true;
            }
            if(blocked){
                A[i][0] = 0;
            }
            else{
                A[i][0] = 1;
            }
                
        }
        
        blocked = false;
        for(int j = 0; j < n; j++){
            if(obstacleGrid[0][j] == 1){
                blocked = true;
            }
            if(blocked){
                A[0][j] = 0;
            }
            else{
                A[0][j] = 1;
            }
        }
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(obstacleGrid[i][j] == 1){
                    A[i][j] = 0;
                }
                else{
                    A[i][j] = A[i - 1][j] + A[i][j - 1];
                }
            }
        }
        
        return A[m - 1][n - 1];
    }
}