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62. Unique Paths && 63 Unique Paths II

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:

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;
};
View Code

 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];        
    }
};

 

posted on 2016-03-16 10:42  鼬与轮回  阅读(177)  评论(0编辑  收藏  举报