LeetCode -- 63 Unique Paths II

LeetCode -- 63 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 and0 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 andn will be at most 100.

解答：基本思想同上一题，运用动态规划的思想 dp[i][j] = dp[i-1][j] + dp[i][j-1]. 本题与上一题的区别在于存在障碍，可把障碍点的dp[i][j]看作0即可，另需要考虑的是当i为0和j为0时存在的障碍点，此种情况，之后运功break即可，后面无需再处理。

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        int m = obstacleGrid.length ;         int n = obstacleGrid[0].length;        int[][] result = new int [m][n];                for(int i = 0; i < m; i++){            if(obstacleGrid[i][0] != 1) result[i][0] = 1;            else break;        }        for(int i = 0; i < n; i++){            if(obstacleGrid[0][i] != 1) result[0][i] = 1;            else break;        }                for(int i = 1; i < m ; i++){            for(int j = 1 ; j < n ; j++){                if(obstacleGrid[i][j] == 1)  result[i][j] = 0;                else result[i][j] = result[i-1][j] + result[i][j-1];            }        }        return result[m-1][n-1];    }}