剑指 Offer 13. 机器人的运动范围
剑指 Offer 13. 机器人的运动范围
难度⭐⭐
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 2：
输入：m = 3, n = 1, k = 0
输出：1
BFS
class Solution:
def movingCount(self, m: int, n: int, k: int) -> int:
grid = [[0]*n for _ in range(m)]
q = [[0,0]]
count = 0
while q:
x,y = q.pop(0)
#如果没有访问过 并且和小于k 就可以进入
if grid[x][y] == 0 and sum(list(map(int,str(x)+str(y)))) <= k:
count += 1
else:
continue
grid[x][y] = 1
for dx,dy in ([1,0],[0,1]):
nx, ny = x+dx, y+dy
if 0 <= nx < m and 0 <= ny <n:
q.append([nx,ny])
return count
import java.util.*;
class Solution {
public int movingCount(int m, int n, int k) {
// 创建队列
Queue<int[]> queue = new LinkedList<int[]>();
int[][] grid = new int[m][n];
for (int i=0; i<m; i++){
for (int j=0;j<n;j++){
grid[i][j] = 0;
}
}
int count=0;
queue.offer(new int[] {0,0});
int[] dx = {0,1};
int[] dy = {1,0};
while (!queue.isEmpty()){
int cor[] = queue.poll();
int x = cor[0], y = cor[1];
if (grid[x][y] == 0 && digitSum(x,y) <= k){
count += 1;
}else{
continue;
}
grid[x][y] = 1;
for (int i=0;i<2;i++){
int nx = x+dx[i];
int ny = y+dy[i];
if (nx>=0 && nx<m && ny>=0 && ny <n){
queue.offer(new int[] {nx,ny});
}
}
}
return count;
}
// 数位求和方法
int digitSum(int x, int y){
int sum = 0;
while (x !=0 || y != 0){
sum = sum + x%10 + y%10;
x = x/10;
y = y/10;
}
return sum;
}
}
DFS
python版一
class Solution:
def movingCount(self, m: int, n: int, k: int) -> int:
def digitSum(x,y):
ans = 0
while x or y:
ans = ans+ x%10 + y%10
x, y = x//10,y//10
return ans
def dfs(x,y):
#递归结束条件
if x<0 or x>=m or y<0 or y>=n or grid[x][y]==1 or digitSum(x,y) > k:
return 0
grid[x][y] = 1
self.count += 1
dfs(x+1,y)
dfs(x,y+1)
grid = [[0]*n for _ in range(m)]
self.count = 0
dfs(0,0)
return self.count
python版二
class Solution:
def movingCount(self, m: int, n: int, k: int) -> int:
def digitSum(x,y):
ans = 0
while x or y:
ans = ans+ x%10 + y%10
x, y = x//10,y//10
return ans
def dfs(x,y):
#递归结束条件
if x<0 or x>=m or y<0 or y>=n or grid[x][y]==1 or digitSum(x,y) > k:
return 0
grid[x][y] = 1
return 1 + dfs(x+1,y) + dfs(x,y+1)
grid = [[0]*n for _ in range(m)]
return dfs(0,0)
java递归版
import java.util.*;
class Solution {
int m,n,k;
boolean[][] grid;
public int movingCount(int m, int n, int k) {
this.m = m; this.n = n; this.k = k;
grid = new boolean[m][n];
return dfs(0,0);
}
int dfs(int x, int y){
//递归结束条件
if(x<0 || x>= m || y<0 || y>=n || grid[x][y]==true || digitSum(x,y) >k){
return 0;
}
grid[x][y] = true;
return 1 + dfs(x+1,y) + dfs(x,y+1);
}
int digitSum(int x, int y){
int sum = 0;
while (x !=0 || y != 0){
sum = sum + x%10 + y%10;
x = x/10;
y = y/10;
}
return sum;
}
}
java栈
import java.util.*;
class Solution {
public int movingCount(int m, int n, int k) {
Stack<int[]> stack = new Stack<int[]>();
stack.push(new int[] {0,0});
int count = 0;
boolean[][] grid = new boolean[m][n];
int[] dx = {1,0};
int[] dy = {0,1};
while(!stack.isEmpty()){
int[] temp = stack.pop();
int x = temp[0];
int y = temp[1];
//满足条件就加一
if (x>=0 && x<m && y>=0 && y <n && grid[x][y] == false && digitSum(x,y) <= k){
count += 1;
}else{
continue;
}
grid[x][y] = true;
for(int i=0;i<2;i++){
int nx = x+dx[i];
int ny = y+dy[i];
stack.push(new int[] {nx,ny});
}
}
return count;
}
int digitSum(int x, int y){
int sum = 0;
while (x !=0 || y != 0){
sum = sum + x%10 + y%10;
x = x/10;
y = y/10;
}
return sum;
}
}
