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

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;

        int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];

        for (int i = 0; i < obstacleGrid.length; i++) {
            if (obstacleGrid[i][0] == 1 || i > 0 && dp[i - 1][0] == 0) {
                dp[i][0] = 0;
            } else {
                dp[i][0] = 1;
            }
        }
        for (int i = 0; i < obstacleGrid[0].length; i++) {
            if (obstacleGrid[0][i] == 1 || i > 0 && dp[0][i - 1] == 0) {
                dp[0][i] = 0;
            } else {
                dp[0][i] = 1;
            }
        }
        for (int i = 1; i < obstacleGrid.length; i++) {
            for (int j = 1; j < obstacleGrid[0].length; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                } else {
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
                }
            }
        }
        return dp[obstacleGrid.length - 1][obstacleGrid[0].length - 1];

    }
}

别人的算法

一维DP

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    int width = obstacleGrid[0].length;
    int[] dp = new int[width];
    dp[0] = 1;
    for (int[] row : obstacleGrid) {
        for (int j = 0; j < width; j++) {
            if (row[j] == 1)
                dp[j] = 0;
            else if (j > 0)
                dp[j] += dp[j - 1];
        }
    }
    return dp[width - 1];
}

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

更多讨论：

https://discuss.leetcode.com/category/71/unique-paths-ii