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 障碍 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.

 
思路:整体思路和Unique Paths一致,就是当a[i][j]=1时,b[i][j]=0,而不是b[i-1][j]+b[i][j-1]。注意初始化a[i][0]和a[0][i]时,只要第一行或者第一列中出现了障碍物,那么就不可能再接着向右走与向下走了。所以要这样初始化:首先初始化b[0][0],如果a[0][0]=1,那么b[0][0]=0,否则为1。后面继续设置a[][]
class Solution {
private:
    int b[101][101];
public:
    int uniquePathsWithObstacles(vector<vector<int>>& a) {
        int i,j;
        int row=a.size();
        int col=a[0].size();
        b[0][0]= a[0][0]==1? 0:1;
        for(i=1;i<row;i++)
            b[i][0]= a[i][0]==1? 0:b[i-1][0];
        for(i=1;i<col;i++)
            b[0][i]= a[0][i]==1? 0:b[0][i-1];
        for(i=1;i<row;i++){
            for(j=1;j<col;j++){
                b[i][j]= a[i][j]==1?  0:b[i-1][j]+b[i][j-1];
            }
        }
        return b[row-1][col-1];
    }
};
原文地址:https://www.cnblogs.com/zhoudayang/p/5013398.html