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;
    }

}