Leetcode 63. Unique Paths II

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.

思路：动态规划题。

一开始边界的初始化很重要，其他的位置，如果是1，则置为0，否则的dp[i][j] = dp[i-1][j] + dp[i][j-1];

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
            return 0;
        int dp[101][101], i, j;
        if(obstacleGrid[0][0] == 0){
            dp[0][0] = 1;
        }
        else{
            return 0;
        }
        
        for(i = 1; i < obstacleGrid.size(); i++){
            if(obstacleGrid[i][0] == 0 && dp[i-1][0] == 1)
                dp[i][0] = 1;
            else
                dp[i][0] = 0;
        }
        
        for(j = 1; j < obstacleGrid[0].size(); j++){
            if(obstacleGrid[0][j] == 0 && dp[0][j-1] == 1)
                dp[0][j] = 1;
            else
                dp[0][j] = 0;
        }
        
        for( i = 1; i < obstacleGrid.size(); i++){
            for(j = 1; j < obstacleGrid[0].size(); j++){
                if(obstacleGrid[i][j] == 0)
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                else
                    dp[i][j] = 0;
            }
        }
        
        return dp[i-1][j-1];
        
    }
};

可以进行代码优化，节省空间。。。以后再说吧