63. 不同路径 II

目录

• 题目描述
• 代码实现

题目描述

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.

代码实现

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {

        int m = obstacleGrid.size();
        if(m == 0)
            return 0;
        int n = obstacleGrid[0].size();
        vector<unsigned> sum(n,0);

        for(int i = 0;i < m;i++)
        {
            for(int j = 0;j < n;j++)
            {   //这个条件可以行使一票否决权
                if(obstacleGrid[i][j] == 1)//流式处理，先不忙操作，先对当前情况进行判断
                    sum[j] = 0;            //再根据判断结果进行对应的操作
                else//下边就是能够到达的情况
                {   
                    if(i == 0 && j == 0)
                        sum[j] = 1 ;
                    else
                    { 
                        unsigned left = 0;//根据递推公式合理的进行初始化
                        if(j !=0)
                            left = sum[j-1];
                        unsigned up = 0;
                        if(i != 0)
                            up = sum[j];
                        sum[j] = left + up;
                    }

                }

            }
        }
        return sum[n-1];       
    }
};