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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：如果地图上有障碍物，此障碍物不能通过 在数组中用1表示障碍物，求从起点到终点有多少种走法。

思路

val[i][0] = 1val[0][j] = 1val[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

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];    }};
    
   

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