• 左上角那个点：有obstacle，则为0；没有obstacle，则为1。
• 最上一排：有obstacle，则为0；没有obstacle，则为左边元素的值。
• 最左边一排：有obstacle，则为0；没有obstacle，则为上边元素的值。
• 其它点：有obstacle，则为0；没有obstacle，则为上边元素和左边元素值得和。

Implementation

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        int right = obstacleGrid[0].length;        int down = obstacleGrid.length;        int[] record = new int[right];        for (int i = 0; i < down; i++) {            for (int j = 0; j < right; j++) {                if (i == 0 && j == 0) {                    record[j] = (obstacleGrid[i][j] == 1)? 0: 1;                }                else if (i == 0) {                    record[j] = (obstacleGrid[i][j] == 1)? 0: record[j - 1];                }                else if (j == 0) {                    record[j] = (obstacleGrid[i][j] == 1)? 0: record[j];                }                else {                    record[j] = (obstacleGrid[i][j] == 1)? 0: record[j] + record[j - 1];                }            }        }        return record[right - 1];    }}