题目:地上有一个m行n列的方格。一个机器人从坐标(0,0)的格子开始移动,它每次可以向左、右、上、下移动一格,但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,但它不能进入方格(35,38),因为3+5+3+8=19,请问该机器人能够到达多少个格子
思路:使用回溯法,判断下一格的数位之和是否小于threshold,小于则加一。
解决方案:
public class Question12 {
public static int movingCount(int threshold, int rows, int cols){
if (threshold < 0 || rows < 1 || cols < 1){
return 0;
}
boolean[] isVisited = new boolean[rows * cols];
int count = movingCountCore(threshold, rows, cols, 0, 0, isVisited);
return count;
}
private static int movingCountCore(int threshold, int rows, int cols, int row, int col, boolean[] isVisited){
int count = 0;
if (check(threshold, rows, cols, row, col, isVisited)){
isVisited[row * cols + col] = true;
count = 1 + movingCountCore(threshold, rows, cols, row, col - 1, isVisited)
+ movingCountCore(threshold, rows, cols, row - 1, col, isVisited)
+ movingCountCore(threshold, rows, cols, row, col + 1, isVisited)
+ movingCountCore(threshold, rows, cols, row + 1, col, isVisited);
}
return count;
}
private static boolean check(int threshold, int rows, int cols, int row, int col, boolean[] isVisited) {
if (row >= 0 && row < rows && col >= 0 && col < cols && getDigitSum(row) + getDigitSum(col) <= threshold && !isVisited[row * cols + col]) {
return true;
}
return false;
}
private static int getDigitSum(int number){
int sum = 0;
while(number > 0){
sum += number % 10;
number /= 10;
}
return sum;
}
public static void main(String[] args) {
System.out.println(movingCount(18, 30,40));
}
}
网友评论