LeetCode Unique Paths II

Description:

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.

Solution:

这道题目和Unique Paths本质上是一样的，只是多了一个额外的限制条件，obstacleGrid数组的限制，那么只要我在初始化和状态表示的时候进行相应的对-1的特判即可。

1.先把所有在obstacleGrid中表示无法通行的点在dp[i][j]中用-1表示

2.初始化中，dp[0][0]如果为-1则返回0；否则为1。然后对于第0行，每一个dp[0][i]如果不为-1，则与前一个dp[0][i-1]相同，同为0或者同为1。第0列也类似。

3.在后续状态变化的过程中，状态的变化方程和前面一样，也是dp[i][j] = dp[i-1][j] + dp[i][j-1]，只不过多了个限制条件，就是只有在dp[i][j-1] != -1的情况下可以加dp[i][j-1]，在dp[i-1][j] != -1的情况下可以加dp[i-1][j]。

注意：这里的初始化需要多考虑考虑。

import java.util.*;public class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int dp[][] = new int[m][n];if (obstacleGrid[0][0] == 1)return 0;elsedp[0][0] = 1;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (obstacleGrid[i][j] == 1)dp[i][j] = -1;for (int i = 1; i < m; i++)if (dp[i][0] != -1) {if (dp[i - 1][0] != -1)dp[i][0] = dp[i - 1][0];}for (int i = 1; i < n; i++)if (dp[0][i] != -1) {if (dp[0][i - 1] != -1)dp[0][i] = dp[0][i - 1];}for (int i = 1; i < m; i++)for (int j = 1; j < n; j++)if (dp[i][j] == 0) {if (dp[i][j - 1] >= 0)dp[i][j] += dp[i][j - 1];if (dp[i - 1][j] >= 0)dp[i][j] += dp[i - 1][j];}if (dp[m - 1][n - 1] == -1)return 0;return dp[m - 1][n - 1];}}