leetcode_c++：Unique Paths II（063）

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.

---

算法

DP
复杂度：O（nm）

同Unique Paths，只是多个障碍物
* 如果是障碍物，则res[i][j]=0
* 否则，res[i][j]=res[i-1][j]+res[i][j-1]

---

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {        if(obstacleGrid.empty() ||obstacleGrid[0].empty())            return 0;        int m=obstacleGrid.size();        int n=obstacleGrid[0].size();        int dp[m][n ];        // 对dp初始化，需要更加obstacGrid的值来确定        dp[0][0]=(obstacleGrid[0][0]==0?1:0);        //我们需要注意m*1和1*n的初始化        for(int i=1;i<m;i++)            dp[i][0]=((dp[i-1][0]==1 && obstacleGrid[i][0]==0)?1:0);        for(int j=1;j<n;j++)            dp[0][j]=((dp[0][j-1]==1 && obstacleGrid[0][j]==0)?1:0);        for(int i=1;i<m;i++)            for(int j=1;j<n;j++){                if(obstacleGrid[i][j]==1)                    dp[i][j]=0;                else                    dp[i][j]=dp[i-1][j]+dp[i][j-1];            }        return dp[m-1][n-1];    }};