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] ]
题目大意
寻求最短路径，从左上走到右下，保证每次只能往左走或往下走（不可以斜着走）。其中数字1是障碍，表示“此路不通”，求总共的路线数
思路
1. 如果没有障碍
val[i][0] = 1
val[0][j] = 1
val[i][j] = val[i-1][j] + val[i][j-1]
2. 有了障碍后
如果obstacle[i][j] = 1
val[i][j] = 1
否则
tmp = obstacle[i-1][j] == 1 ? 0 : val[i-1][j]
tmp = obstacle[i][j-1] == 1 ? tmp : tmp + val[i-1][j-1]
val[i][j] = tmp
参考代码
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { int row = obstacleGrid.size(); int col = obstacleGrid[0].size(); int token = 1; int val[row][col]; for (int j = 0; j < col; ++j) { if(obstacleGrid[0][j] == 1) token = 0; val[0][j] = token; } token = 1; for (int i = 0; i < row; ++i) { if(obstacleGrid[i][0] == 1) token = 0; val[i][0] = token; } for (int i = 1; i < row; ++i) { for(int j = 1; j < col; ++j) { if (obstacleGrid[i][j] == 1) val[i][j] = 0; else { int tmp = obstacleGrid[i-1][j] == 1 ? 0 :val[i-1][j]; tmp = obstacleGrid[i][j-1] == 1 ? tmp : tmp + val[i][j-1]; val[i][j] = tmp; } } } return val[row-1][col-1]; } };
本文转自jihite博客园博客，原文链接：http://www.cnblogs.com/kaituorensheng/p/3806896.html，如需转载请自行联系原作者
