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?
分析
该题和矩阵中的路径又有些相似,都是回溯,都是上下左右"走路"问题,代码都做了相关注释
public class Solution {
public int movingCount(int threshold, int rows, int cols)
{
// 参数有效性检查
if (threshold <= 0 || rows <= 0 || cols <= 0) {
return 0;
}
boolean[] visited = new boolean[rows * cols];//默认初始化为false
return movingCountCore(threshold, rows, cols, 0, 0, visited);
}
private static int movingCountCore(int threshold, int rows, int cols, int row, int col, boolean[] visited) {
int count = 0;
//判断是否能进入这个格子,即未被访问且满足题限制
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;//走过的格子数,即能够达到的格子
}
//判断
private static boolean check(int threshold, int rows, int cols, int row, int col, boolean[] visited) {
if (row >= 0 && row < rows && col >= 0 && col < cols && !visited[row * cols + col] && getSum(row, col) <= threshold) {
return true;
}
return false;
}
//求数位之和
private static int getSum(int row, int col) {
int sum = 0;
while (row > 0) {
sum += row % 10;
row /= 10;
}
while (col > 0) {
sum += col % 10;
col /= 10;
}
return sum;
}
}