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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.
思路：
关于存储空间的优化，可以看姐妹题：Leetcode 62. Unique Paths
1. 如果地图的左上角定格是障碍，则输出为0；
2. 对于第0行的第j(0<j<n)个，cur[0]=1：
2.1 obstacleGrid[0][j-1]==1（等价于cur[j-1]==0），则cur[j]=0
2.2 obstacleGrid[0][j-1]==0（等价于cur[j-1]==1），则cur[j]=1-obstacleGrid[0][j]
3. 对于第i行(0<i<m)：
3.1 对于第0个：
3.1.1 obstacleGrid[i-1][0]==1，cur[0]=0
3.1.2 obstacleGrid[i-1][0]==0，cur[0]=1
3.2 对于第j(0<j<n)个：
3.2.1 obstacleGrid[i][j]==1，cur[j]=0
3.2.2 obstacleGrid[i][j]==0，cur[j]=cur[j-1]+cur[j]
代码：
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 int m=obstacleGrid.size(); 5 int n=obstacleGrid[0].size(); 6 vector<int> cur(n); 7 int i,j; 8 if(obstacleGrid[0][0]){ 9 return 0; 10 } 11 cur[0]=1; 12 for(j=1;j<n;j++){ 13 cur[j]=cur[j-1]?1-obstacleGrid[0][j]:0; 14 } 15 for(i=1;i<m;i++){ 16 cur[0]=cur[0]?1-obstacleGrid[i][0]:0; 17 for(j=1;j<n;j++){ 18 cur[j]=obstacleGrid[i][j]?0:cur[j]+cur[j-1]; 19 } 20 } 21 return cur[n-1]; 22 } 23 };
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原文地址：http://www.cnblogs.com/Deribs4/p/5651709.html
