地上有一个m行和n列的方格。一个机器人从坐标0,0的格子开始移动,每一次只能向左,右,上,下四个方向移动一格,但是不能进入行坐标和列坐标的数位之和大于k的格子。 例如,当k为18时,机器人能够进入方格(35,37),因为3+5+3+7 = 18。但是,它不能进入方格(35,38),因为3+5+3+8 = 19。请问该机器人能够达到多少个格子?
基本上就是列好坐标系,然后挨个占位置,占到位置的标1,条件不允许的就是0
# -*- coding:utf-8 -*-
class Solution:
def movingCount(self, threshold, rows, cols):
index = [0] * (rows * cols)
i = 0
j = 0
return self.moving(threshold, i, j, rows, cols, index)
def moving(self, threshold, i, j, rows, cols, index):
count = 0
if i >= 0 and j >= 0 and i<rows and j<cols and index[i*cols+j] == 0 and self.axis_sum(i)+self.axis_sum(j) <= threshold:
index[i * cols + j] = 1
count = 1 + \
self.moving(threshold, i, j + 1, rows, cols, index) + \
self.moving(threshold, i - 1, j, rows, cols, index) + \
self.moving(threshold, i + 1, j, rows, cols, index) + \
self.moving(threshold, i, j - 1, rows, cols, index)
#上面一行就是朝四个方向拓展,大概是我一辈子也没法自己想出来的逻辑 = =
return count
def axis_sum(self, k):
m = 0
while k > 0:
m = m + k%10
k = round(k/10-0.5)
return m
# write code here
#include
#include
#include
using namespace std;
class Solution {
public:
int movingCount(int threshold, int rows, int cols)
{
vector<int>vec(rows*cols,0);
return moving(threshold, vec, 0, 0, rows, cols);
}
int moving(int threshold, vector<int>&vec, int i, int j, int rows, int cols)
{
int count=0;
if(i>=0 && i<rows && j>=0 && j<cols && vec[i*cols+j]==0 && getsum(i)+getsum(j)<=threshold) //vec就是用来占(数)位置的,占过的位置变成1,然后继续占还是0的位置
{
vec[i*cols+j]=1;
count=1+moving(threshold, vec, i+1, j, rows, cols)+
moving(threshold, vec, i-1, j, rows, cols)+
moving(threshold, vec, i, j+1, rows, cols)+
moving(threshold, vec, i, j-1, rows, cols);
}
return count;
}
int getsum(int x)
{
int sum=0;
while(x>0)
{
sum=sum+x%10;
x=x/10;
}
return sum;
}
};
int main()
{
int threshold=5;
int cols=10;
int rows=10;
Solution a;
cout << "Count : " << a.movingCount(threshold,rows,cols) << endl;
return 0 ;
}
虽然去看了一下self的理解,但感觉还没有弄明白,哪天专门写一个整理的吧