题目描述
地上有一个m行和n列的方格。一个机器人从座标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,但是不能进入行座标和列座标的数位之和大于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:将地图全部置1,遍历能够到达的点,将遍历的点置0并令计数+1.这个思路在找前后左右相连的点很有用,比如leetcode中的海岛个数问题/最大海岛问题都可以用这种方法来求解。
python
# -*- coding:utf-8 -*-
class Solution:
def __init__(self):
self.count = 0
def movingCount(self, threshold, rows, cols):
# write code here
arr = [[1 for i in range(cols)] for j in range(rows)]
self.findway(arr, 0, 0, threshold)
return self.count
def findway(self, arr, i, j, k):
if i < 0 or j < 0 or i >= len(arr) or j >= len(arr[0]):
return
tmpi = list(map(int, list(str(i))))
tmpj = list(map(int, list(str(j))))
if sum(tmpi) + sum(tmpj) > k or arr[i][j] != 1:
return
arr[i][j] = 0
self.count += 1
self.findway(arr, i + 1, j, k)
self.findway(arr, i, j + 1, k)
c++
class Solution {
public:
int movingCount(int threshold, int rows, int cols)
{
bool *flag = new bool[rows * cols];
for(int i = 0; i < rows * cols; i++)
flag[i] = false;
int count = moving(threshold, rows, cols, 0, 0, flag);
return count;
}
int moving(int threshold, int rows, int cols, int i, int j, bool* flag)
{
int count = 0;
if(i >= 0 && i < rows && j >= 0 && j < cols && getsum(i) + getsum(j) <= threshold && flag[i * cols + j] == false)
{
flag[i * cols + j] = true;
count =1 + moving(threshold, rows, cols, i + 1, j, flag)
+ moving(threshold, rows, cols, i - 1, j, flag)
+ moving(threshold, rows, cols, i , j - 1, flag)
+ moving(threshold, rows, cols, i, j + 1, flag);
}
return count;
}
int getsum(int num)
{
int sum = 0;
while(num)
{
sum += num % 10;
num /= 10;
}
return sum;
}
};