题目描述
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?
代码
代码语言:javascript复制//
// Created by jiangxingqi on 2019/4/16.
//
#include <iostream>
class Solution {
public:
int movingCount(int threshold, int rows, int cols)
{
bool* visited = new bool[rows*cols];
memset(visited, 0, rows*cols);
int count = movingCountCore(threshold, rows, cols, 0, 0, visited);
delete[]visited;
return count;
}
int movingCountCore(int threshold, int rows, int cols, int row, int col, bool* visited)
{
int count = 0;
if (check(threshold, rows, cols, row, col, visited))
{
visited[row*cols col] = true;
count = 1 movingCountCore(threshold, rows, cols, row, col - 1, visited)
movingCountCore(threshold, rows, cols, row - 1, col, visited)
movingCountCore(threshold, rows, cols, row, col 1, visited)
movingCountCore(threshold, rows, cols, row 1, col, visited);
}
return count;
}
/*该函数检查坐标为(row,col)的方格能够进入*/
bool check(int threshold, int rows, int cols, int row, int col, bool*visited)
{
if (row >= 0 && row < rows&&col >= 0 && col < cols
&&getDigitSum(row) getDigitSum(col) <= threshold
&& !visited[row*cols col])
return true;
return false;
}
/*计算一个数的所有位数之和*/
int getDigitSum(int number)
{
int sum = 0;
while (number > 0)
{
sum = number % 10;
number = number / 10;
}
return sum;
}
};
int main() {
Solution *ss=new Solution();
int count = ss->movingCount(4,10,10);
std::cout << count << std::endl;
return 0;
}