[LeetCode]63 不同的路径总数之二

Unique Paths II(不同的路径总数之二)

【难度：Medium】
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.

LeetCode 62题的扩展，为机器人增加了障碍物，0表示空地，1表示障碍，同样求从左上角到右下角不同的走法总数。

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解题思路

在62题的基础上解决这个问题，但需要考虑的是，机器人不能简单地向下或向右走，要判断是否有障碍物，因此左边界和上边界的初始化也要根据上一个点的状态来设置，同样使用动态规划来完成。

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c++代码如下：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        if (obstacleGrid.empty())            return 0;        //入口有障碍的话，直接返回        if (obstacleGrid[0][0] == 1)            return 0;        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector<vector<int>> path(m,vector<int>(n,0));        //入口点走法只有1种。        path[0][0] = 1;        for (int i = 1; i < m; i++) {            //左边界，如果上一点可行并且当前点没有障碍物，那么该点可走            if (path[i-1][0] != 0 && obstacleGrid[i][0] != 1)                path[i][0] = 1;        }        for (int i = 1; i < n; i++) {            //上边界与左边界情况同理            if (path[0][i-1] != 0 && obstacleGrid[0][i] != 1)                path[0][i] = 1;        }        for (int i = 1; i < m; i++)            for (int j = 1; j < n; j++)                //动态规划，当前点无障碍物则与62题处理方法一致                if (obstacleGrid[i][j] != 1)                    path[i][j] = path[i-1][j]+path[i][j-1];        return path[m-1][n-1];    }};