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.
寻求最短路径，从左上走到右下，保证每次只能往左走或往下走（不可以斜着走）。其中数字1是障碍，表示“此路不通”，求总共的路线数。
第一种二维数组
用一个二维数组来表示前者的路径 核心就是这个，如果不等于1，我们就找到前者的路径相加。
if (obstacleGrid[i][j] == 1) { continue; } else { int tmp = obstacleGrid[i - 1][j] == 1 ? 0 : val[i - 1][j]; tmp = obstacleGrid[i][j - 1] == 1 ? tmp : tmp + val[i][j - 1]; val[i][j] = tmp; } public int uniquePathsWithObstacles1(int[][] obstacleGrid) { if (obstacleGrid[0][0] == 1) return 0; int m = obstacleGrid.length; int n = obstacleGrid[0].length; int[][] val = new int[m][n]; val[0][0] = 1; for (int i = 1; i < m; i++) if (obstacleGrid[i][0] != 1 && val[i - 1][0] != 0) val[i][0] = 1; for (int i = 1; i < n; i++) if (obstacleGrid[0][i] != 1 && val[0][i - 1] != 0) val[0][i] = 1; for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { if (obstacleGrid[i][j] == 1) { continue; } else { int tmp = obstacleGrid[i - 1][j] == 1 ? 0 : val[i - 1][j]; tmp = obstacleGrid[i][j - 1] == 1 ? tmp : tmp + val[i][j - 1]; val[i][j] = tmp; } } } return val[m - 1][n - 1]; }第二种一维数组
其实一维数组足以表示前者的路径，因为一维数组左边是你更新过的，右边是没更新，没更新的相当于上一排，也就是上一排的来路加上左边的来路之和就是现在的来路。（解释好混乱，但我是这样想就理解了）
public int uniquePathsWithObstacles(int[][] obstacleGrid) { if (obstacleGrid[0][0] == 1) return 0; int m = obstacleGrid.length; int n = obstacleGrid[0].length; int[] step = new int[n]; step[0] = 1; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) { if (obstacleGrid[i][j] == 1) step[j] = 0; else if (j > 0) step[j] += step[j - 1]; } return step[n - 1]; }