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]]

题目大意

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

思路

   1. 如果没有障碍

val[i][0] = 1 val[0][j] = 1 val[i][j] = val[i-1][j] + val[i][j-1] 2. 有了障碍后 如果obstacle[i][j] = 1      val[i][j] = 1 否则      tmp = obstacle[i-1][j] == 1 ? 0 : val[i-1][j]      tmp = obstacle[i][j-1] == 1 ? tmp : tmp + val[i-1][j-1] 　　　val[i][j] = tmp 参考代码

class Solution {public:    int uniquePathsWithObstacles(vector
     
      
        > &obstacleGrid) {        int row = obstacleGrid.size();        int col = obstacleGrid[0].size();        int token = 1;        int val[row][col];        for (int j = 0; j < col; ++j)        {            if(obstacleGrid[0][j] == 1)                token = 0;            val[0][j] = token;                    }        token = 1;        for (int i = 0; i < row; ++i)        {            if(obstacleGrid[i][0] == 1)                token = 0;            val[i][0] = token;        }        for (int i = 1; i < row; ++i)        {            for(int j = 1; j < col; ++j)            {                if (obstacleGrid[i][j] == 1)                    val[i][j] = 0;                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[row-1][col-1];    }};