【Leetcode长征系列】Unique Paths II

原题：

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：一开始的想法太复杂了，想的是先遍历一遍如果遇到为1则置为INT_MIN，之后在扫矩阵的时候遇到INT_MIN就忽略加0；但其实只需要遍历一遍就够了，在遇到1的时候置0即可。还有问题就是，if语句的分情况要完整，不然很容易出问题。

代码：

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。