LeetCode Unique Paths II

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.

题意：

此题是Unique Paths的一个变种，就是在二维数组中增加一些特殊的障碍，但是其实归根到底都是DP(动态规划)的类型。考虑几种特殊情况就行了，首先这里增加了一个obstacleGrid数组，这个数组中存储有每个位置上的值，1代表有障碍，0代表没有障碍；那么我另外又开了一个二维数组用来保存走到每一个位置的种树。但是这里要考虑的是，在第一行和第一列中，如果一旦一个位置出现了1，那么后面的所有格子都将是0，这里要特别注意。另外格子和之前考虑的一样，都是左边和上边的那两个格子的值相加。

public static int uniquePathsWithObstacles(int[][] obstacleGrid){int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] nums = new int[m][n];if(obstacleGrid[0][0] == 1)return 0;nums[0][0] = 1;for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(i == 0 && j > 0){if(obstacleGrid[i][j] == 0 && obstacleGrid[i][j - 1] != 1 && nums[i][j-1] != 0)nums[i][j] = 1;else if(obstacleGrid[i][j] == 0 && obstacleGrid[i][j - 1] == 1)nums[i][j] = 0;else if(obstacleGrid[i][j] == 1)nums[i][j] = 0;}if(i > 0 && j == 0){if(obstacleGrid[i][j] == 0 && obstacleGrid[i-1][j] != 1 && nums[i-1][j] != 0)nums[i][j] = 1;else if(obstacleGrid[i][j] == 0 && obstacleGrid[i-1][j] == 1)nums[i][j] = 0;else if(obstacleGrid[i][j] == 1)nums[i][j] = 0;}if(i > 0 && j > 0){if(obstacleGrid[i][j] == 1)nums[i][j] = 0;else if(obstacleGrid[i][j] != 1)nums[i][j] = nums[i-1][j] + nums[i][j-1];}}}return nums[m-1][n-1];}