LeetCode | Unique Paths II
2013-12-23 18:25
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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
1and
0respectively
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:
只需要在/article/1382539.html的基础上增加一个判断条件。
代码:
方法1:class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { int m=obstacleGrid.size(); int n = obstacleGrid[0].size(); int ** dp=new int*[m]; for(int i=0;i<m;i++) { dp[i] = new int ; } int step = m+n-1; int cur=1; dp[0][0]=(obstacleGrid[0][0]==1?0:1); while(cur<step) { for(int i=0;i<=cur;i++) { if(i>=m) { continue; } int j=cur-i; if(j>=n) { continue; } if(obstacleGrid[i][j]==1) { dp[i][j]=0; } else { dp[i][j]=0; if(i>0) { dp[i][j]+=dp[i-1][j]; } if(j>0) { dp[i][j]+=dp[i][j-1]; } } } cur++; }; return dp[m-1][n-1]; } };
方法2:
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0){ return 0; } int m = obstacleGrid.size(); int n = obstacleGrid[0].size(); int** dp = new int*[m]; for(int i = 0; i < m; i++){ dp[i] = new int ; for(int j = 0; j < n; j++){ dp[i][j] = 0; } } dp[0][0] = (obstacleGrid[0][0] == 0) ? 1 : 0; for(int i = 1; i < m; i++){ dp[i][0] = (obstacleGrid[i][0] == 0) ? dp[i-1][0] : 0; } for(int i = 1; i < n; i++){ dp[0][i] = (obstacleGrid[0][i] == 0) ? dp[0][i-1] : 0; } for(int i = 1; i < m; i++){ for(int j = 1; j < n; j++){ dp[i][j] = dp[i-1][j] + dp[i][j-1]; if(obstacleGrid[i][j] == 1){ dp[i][j] = 0; } } } return dp[m-1][n-1]; } };
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