原题地址:https://leetcode-cn.com/problems/ji-qi-ren-de-yun-dong-fan-wei-lcof/
题目描述:
地上有一个m行n列的方格,从坐标 [0,0] 到坐标 [m-1,n-1] 。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(不能移动到方格外),也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?
示例 1:
输入:m = 2, n = 3, k = 1
输出:3
示例 1:
输入:m = 3, n = 1, k = 0
输出:1
提示:
1 <= n,m <= 100
0 <= k <= 20
解题方案:
可以用DFS和BFS进行遍历,要注意的是遍历的过程只会往右下方走,所有左上不需要考虑。
代码:
DFS(100%)
class Solution {
int m;
int n;
int k;
int count;
int[][] visited;
public int movingCount(int m, int n, int k) {
if(m == 0 || n == 0) return 0;
visited = new int[m][n];
this.m = m;
this.n = n;
this.k = k;
count = 0;
DFS(0, 0);
return count;
}
public void DFS(int i, int j)
{
if(i < 0 || i >= m || j < 0 || j >= n || visited[i][j] == 1) return;
int x = i, y = j;
int sum = 0;
while(x != 0){sum += x % 10; x /= 10;}
while(y != 0){sum += y % 10; y /= 10;}
if(sum <= k)
{
visited[i][j] = 1;
count ++;
}
else
return;
DFS(i + 1, j);
DFS(i, j + 1);
}
}
BFS
class Solution {
public int movingCount(int m, int n, int k) {
if(m == 0 || n == 0) return 0;
Queue<Integer> q = new LinkedList<>();
int[][] visited = new int[m][n];
q.add(0); q.add(0);
visited[0][0] = 1;
int count = 0;
int i, j;
while(!q.isEmpty())
{
i = q.remove();
j = q.remove();
count ++;
if(i + 1 < m && visited[i + 1][j] == 0)
{
int sum = 0;
int x = i + 1, y = j;
while(x != 0){ sum += x % 10; x /= 10;}
while(y != 0){ sum += y % 10; y /= 10;}
if(sum <= k)
{
q.add(i + 1); q.add(j);
visited[i + 1][j] = 1;
}
}
if(j + 1 < n && visited[i][j + 1] == 0)
{
int sum = 0;
int x = i, y = j + 1;
while(x != 0){ sum += x % 10; x /= 10;}
while(y != 0){ sum += y % 10; y /= 10;}
if(sum <= k)
{
q.add(i); q.add(j + 1);
visited[i][j + 1] = 1;
}
}
}
return count;
}
}