【Leetcode長征系列】Unique Paths II

原題：

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：一開始的想法太複雜了，想的是先遍歷一遍如果遇到爲1則置爲INT_MIN，之後在掃矩陣的時候遇到INT_MIN就忽略加0；但其實只需要遍歷一遍就夠了，在遇到1的時候置0即可。還有問題就是，if語句的分情況要完整，不然很容易出問題。

代碼：

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

AC。