【Unique Paths II】cpp
题目:
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:
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) { const int m = obstacleGrid.size(); const int n = obstacleGrid[0].size(); vector<vector<int> > cache(m+1,vector<int>(n+1,0)); return Solution::dfs(m, n, cache, obstacleGrid); } static int dfs( int i, int j, vector<vector<int> >& cache, vector<vector<int> >& obstacleGrid ) { if ( i<1 || j<1 || obstacleGrid[i-1][j-1]==1 ) return 0; if ( i==1 && j==1 ) return 1; return Solution::getDFS(i-1, j, obstacleGrid, cache) + Solution::getDFS(i, j-1, obstacleGrid, cache); } static int getDFS(int i, int j, vector<vector<int> >& obstacleGrid, vector<vector<int> >& cache) { if ( cache[i][j]>0 ){ return cache[i][j]; } else{ return cache[i][j] = Solution::dfs(i, j, cache, obstacleGrid); } } };
tips:
上述的代码采用了深搜+缓存(“备忘录”)法。照比Unique Paths多了一个判断条件,如果obstacleGrid上该位置为1则直接返回0。
有个细节上的技巧:
obstacleGrid是题目给出的定义,下标[m-1][n-1]
cache是自定义的缓存数组,下标[m][n]
同样的位置,cache的坐标比obstacleGrid大1;因此,只要深搜过程中保证了cache的坐标在合理范围内,无论是横坐标-1还是纵坐标-1,映射到cache的坐标中总不会越界。
上述代码的效率并不高,但是比较容易处理各种case。再尝试动态规划的解法。
==========================================
在Unique Paths的基础上,用动规又把Unqiue Paths II写了一遍。
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) { const int m = obstacleGrid.size(); const int n = obstacleGrid[0].size(); vector<int> dp(n, 0); if ( obstacleGrid[0][0]==1 ) return 0; dp[0] = 1; for ( size_t i = 0; i < m; ++i ) { dp[0] = obstacleGrid[i][0]==1 ? 0 : dp[0]; for ( size_t j =1; j < n; ++j ) { dp[j] = obstacleGrid[i][j]==1 ? 0 : dp[j-1] + dp[j]; } } return dp[n-1]; } };
tips:
沿用了滚动数组的技巧,额外空间缩减为O(n),代码的效率也提升了。
==========================================
第二次过这道题,用dp过的。
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { if (obstacleGrid.empty()) return 0; const int m = obstacleGrid.size(); const int n = obstacleGrid[0].size(); int dp[m][n]; fill_n(&dp[0][0], m*n, 0); for ( int i=0; i<n; ++i ) { if ( obstacleGrid[0][i]==0 ) { dp[0][i]=1; } else { break; } } for ( int i=0; i<m; ++i ) { if ( obstacleGrid[i][0]==0 ) { dp[i][0]=1; } else { break; } } for ( int i=1; i<m; ++i ) { for ( int j=1; j<n; ++j ) { if ( obstacleGrid[i][j]==0 ) { dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } } return dp[m-1][n-1]; } };