题目描述

写一个Java版的dfs,这道题也可以用暴力解决

样例

class Solution {
     public int movingCount(int threshold, int rows, int cols)
    {
        boolean visit[][]=new boolean[rows][cols];
        return dfs(threshold,0,0,rows,cols,visit);
    }
    public int dfs(int k,int m,int n,int row,int cols,boolean visit[][]){
             if(m>=row||m<0)return 0;
             if(n>=cols||n<0)return 0;
             if(visit[m][n])return 0;
             if(get_number(m, n)>k)return 0;
             visit[m][n]=true;
            return 1+dfs(k, m-1, n, row, cols, visit)+
                    dfs(k, m+1, n, row, cols, visit)+
                    dfs(k, m, n-1, row, cols, visit)+
                    dfs(k, m, n+1, row, cols, visit);
    }
    public int get_number(int m,int n){
        int total=0;
        while(m!=0){
            total+=m%10;
            m=m/10;
        }
        while(n!=0){
            total+=n%10;
            n=n/10;
        }
        return total;
    }
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla