[leetcode]63. Unique Paths II

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.

分析：和前一题差不多的思路，只是遇到1的时候需要跳过去，这一次刷题写的代码更简洁一些，但是卡在了中间一个公式的符号优先级上，事实证明，凡是遇到自己不确定优先级的式子就按照需要加上括号，至少不会出错。

代码：

1.

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

2.

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