1. 题目

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.

2. 思路

第一遍用递归的方式来计算，当矩阵太大了递归太多超时了。

第二遍改为迭代方案，从右下往左上走，保留每个点[i,j]走到底的路径数量，就可以通过加和得到下一步的大小。复杂度O(m*n)。

3. 代码

耗时：6ms

class Solution {public:    int uniquePathsWithObstacles(vector
    
     
      >& obstacleGrid) {        //return upwo(obstacleGrid, 0, 0);        return upwo(obstacleGrid);    }        // 迭代的方案    // 多加一行一列, 填为0，方便迭代计算. 为了简化最右下的点，将右下的右边点设置为1    int upwo(vector
      
       
        >& og) {        int row = og.size();        if (row == 0) { return 0; }        int col = og[0].size();        if (col == 0) { return 0; }        if (og[0][0] == 1 || og[row-1][col-1] == 1) { return 0; }                int way[row+1][col+1];        for (int i = 0; i < row + 1; i++) {            way[i][col] = 0;        }        for (int j = 0; j < col + 1; j++) {            way[row][j] = 0;        }        way[row-1][col] = 1;        for (int i = row-1; i >= 0; i--) {            for (int j = col-1; j >= 0; j--) {                if (og[i][j] == 1) {                    way[i][j] = 0;                } else {                    way[i][j] = way[i+1][j] + way[i][j+1];                }            }        }        return way[0][0];    }        // 超时的方案    // 用递归的方式来计算    // 每次可以向右和相下两种方向，如果某个方向是1，则此方向的种类是0    int upwo(vector
        
         
          >& og, int i, int j) { if (og[i][j] == 1) { return 0; } int row = og.size(); if (row == 0) { return 0; } int col = og[0].size(); if (col == 0) { return 0; } if (i == row - 1 && j == col - 1) { return 1; } if (i == row - 1) { return upwo(og, i, j + 1); } if (j == col - 1) { return upwo(og, i + 1, j); } return upwo(og, i, j + 1) + upwo(og, i + 1, j); }};