【问题】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?
【思路】这道题目我们需要花很多心思来确定递归函数的结束条件:即每个格子能否被访问到!最关键的两个条件是,首先没有标记访问,其次是其坐标的数位之和等于k。如果全部满足,则当前状态等于以上所有子状态的解+1。
class Solution { public: int movingCount(int threshold, int rows, int cols) { if(threshold < 0 || rows < 1 || cols < 1) return false; 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 + 1, col, visited) + movingCountCore(threshold, rows, cols, row - 1, col, visited) + movingCountCore(threshold, rows, cols, row, col + 1, visited) + movingCountCore(threshold, rows, cols, row, col - 1, visited); } return count; } bool check(int threshold, int rows, int cols, int row, int col, bool* visited) { if(row >=0 && row < rows && col >= 0 && col < cols && !visited[row*cols+col] && getNum(row)+getNum(col) <= threshold) return true; return false; } int getNum(int number) { int sum = 0; while(number) { sum += number % 10; number /= 10; } return sum; } };
原文:https://www.cnblogs.com/zhudingtop/p/11494654.html