标签：style blog io ar color sp for on div
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.
思路：跟Unique Paths差不多，就是把有障碍的地方方法数变成0，注意左上角为障碍的情况
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { if(obstacleGrid.empty()) { return 0; } vector<vector<int>> ways(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(),1)); if(obstacleGrid[0][0] == 1) { ways[0][0] = 0; } for(int c = 1; c < obstacleGrid[0].size(); c++) { ways[0][c] = (obstacleGrid[0][c] == 1) ? 0 : ways[0][c - 1]; } for(int r = 1; r < obstacleGrid.size(); r++) { ways[r][0] = (obstacleGrid[r][0] == 1) ? 0 : ways[r - 1][0]; } for(int i = 1; i < obstacleGrid.size(); i++) { for(int j = 1; j < obstacleGrid[0].size(); j++) { ways[i][j] = (obstacleGrid[i][j] == 1) ? 0 : ways[i - 1][j] + ways[i][j - 1]; } } return ways[obstacleGrid.size() - 1][obstacleGrid[0].size() - 1]; } };
【leetcode】 Unique Path ||(easy)
标签：style blog io ar color sp for on div
原文地址：http://www.cnblogs.com/dplearning/p/4146967.html
