leetCode 63.Unique Paths II (唯一路径II) 解题思路和方法

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.

思路：这题理解题意之后并不比上一题麻烦，只是我在刚开始的时候，没有完全想对思路。

在给第一行和第一列赋值时，如果有障碍了，则后面的则全部赋值为0，表示通不过。

具体代码如下：

public class Solution {    public int uniquePathsWithObstacles(int[][] f) {        if(f.length == 0 || f[0].length == 0 || f[0][0] > 0)            return 0;        //其余情况        for(int i = 0; i < f.length; i++){        if(f[i][0] > 0){//如果有障碍，后面全部为0        while(i < f.length){        f[i++][0] = 0;        }        }else{        f[i][0] = 1;//没有障碍赋值为1        }        }        for(int i = 1; i < f[0].length; i++){//从i=1开始，因为f[0][0]值已经改变        if(f[0][i] > 0){//如果有障碍，后面全部为0        while(i < f[0].length){        f[0][i++] = 0;        }        }else{        f[0][i] = 1;        }        }        //循环走到最终点        for(int i = 1; i < f.length; i++)            for(int j = 1; j < f[0].length; j++){                if(f[i][j] > 0)                    f[i][j] = 0;//有障碍，设置0，表示不通                else                    f[i][j] = f[i-1][j] + f[i][j-1];            }        return f[f.length-1][f[0].length-1];    }}