https://leetcode.com/problems/unique-paths/
这道题,不利用动态规划基本上规模变大会运行超时,下面自己写得这段代码,直接暴力破解,只能应付小规模的情形,当23*12时就超时了:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: // Solution():dp1(m,vector<int>(n,-1)),dp2(m,vector<int>(n,-1)){ // } int uniquePaths(int m, int n) { helper(1,1,m,n,dp1,dp2); return res; } void helper(int row,int col,int m,int n,vector<vector<int>>& dp1,vector<vector<int>>& dp2){ if(row==m && col==n){ res++; return; } if(row<m&&col<n){ helper(row+1,col,m,n,dp1,dp2); helper(row,col+1,m,n,dp1,dp2); } else if(row==m && col<n){ helper(row,col+1,m,n,dp1,dp2); } else if(row<m && col==n){ helper(row+1,col,m,n,dp1,dp2); } } private: vector<vector<int>> dp1; vector<vector<int>> dp2; int res; };
利用动态规划来做,储存一个二维数组:vector<vector<int>> dp,用来保存每一个位置的走法总数:
则dp[i][j]=dp[i-1][j]+dp[i][j-1],代码如下:
class Solution { public: int uniquePaths(int m, int n) { vector<vector<int>> dp(m+1,vector<int>(n+1,-1)); helper(1,1,m,n,dp); return dp[m][n]; } void helper(int row,int col,int m,int n,vector<vector<int>>& dp){ for(int i=1;i<=m;i++){ for(int j=1;j<=n;j++){ if(i==1||1==j) dp[i][j]=1; else{ dp[i][j]=dp[i-1][j]+dp[i][j-1]; } } } } };
63 Unique Paths II https://leetcode.com/problems/unique-paths-ii/
递归公式仍旧是上一个,不过遇到障碍要作相应处理(置为零)
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int row=obstacleGrid.size(); int col=obstacleGrid[0].size(); vector<vector<int>> dp(row,vector<int>(col,-1)); for(int i=0;i<row;i++){ for(int j=0;j<col;j++){ if(obstacleGrid[i][j]==1){ dp[i][j]=0; } else{ if(i==0&&0==j){ dp[i][j]=1; } else if(0==i && j>0){ dp[i][j]=dp[i][j-1]; } else if(j==0&& i>0){ dp[i][j]=dp[i-1][j]; } else{ dp[i][j]=dp[i-1][j]+dp[i][j-1]; } } } } return dp[row-1][col-1]; } };