63. Unique Paths II

题目：

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.

链接：https://leetcode.com/problems/unique-paths-ii/#/description

4/26/2017

1ms, 19%

跟unique paths差不多，区别是，如果是障碍那么dp为0，并且初始化边界的时候也要注意这个条件。

注意的问题：

1. 判断非法输入

2. 在初始化第一行和第一列的时候就要想到有obstacle，所以要判断是否当前为障碍，以及如果之前一个dp是0，现在的也应该是0

3. 注意typo，不要随便copy & paste，比如第8行开始是obstacleGrid，后面是dp

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;
 4 
 5         int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];
 6 
 7         for (int i = 0; i < obstacleGrid.length; i++) {
 8             if (obstacleGrid[i][0] == 1 || i > 0 && dp[i - 1][0] == 0) {
 9                 dp[i][0] = 0;
10             } else {
11                 dp[i][0] = 1;
12             }
13         }
14         for (int i = 0; i < obstacleGrid[0].length; i++) {
15             if (obstacleGrid[0][i] == 1 || i > 0 && dp[0][i - 1] == 0) {
16                 dp[0][i] = 0;
17             } else {
18                 dp[0][i] = 1;
19             }
20         }
21         for (int i = 1; i < obstacleGrid.length; i++) {
22             for (int j = 1; j < obstacleGrid[0].length; j++) {
23                 if (obstacleGrid[i][j] == 1) {
24                     dp[i][j] = 0;
25                 } else {
26                     dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
27                 }
28             }
29         }
30         return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
31 
32     }
33 }

别人的算法

一维DP

 1 public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 2     int width = obstacleGrid[0].length;
 3     int[] dp = new int[width];
 4     dp[0] = 1;
 5     for (int[] row : obstacleGrid) {
 6         for (int j = 0; j < width; j++) {
 7             if (row[j] == 1)
 8                 dp[j] = 0;
 9             else if (j > 0)
10                 dp[j] += dp[j - 1];
11         }
12     }
13     return dp[width - 1];
14 }

https://discuss.leetcode.com/topic/10974/short-java-solution