zoukankan      html  css  js  c++  java
  • 剑指 Offer 13. 机器人的运动范围

    剑指 Offer 13. 机器人的运动范围

    地址:剑指 Offer 13. 机器人的运动范围

    问题描述:

    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 2:

    输入:m = 3, n = 1, k = 0
    输出:1
    提示:

    1 <= n,m <= 100
    0 <= k <= 20

    //DFS
    object Solution {
        def movingCount(m: Int, n: Int, k: Int): Int = {
            val visited = Array.fill(m+1, n+1)(false)
            return dfs(visited, m, n, k, 0, 0) 
        }
    
        def dfs(visited: Array[Array[Boolean]], m: Int, n: Int, k: Int, i: Int, j: Int): Int = {
            if (i < 0 || i >= m || j < 0 || j >= n || visited(i)(j) == true || sum(i) + sum(j) > k) return 0
            visited(i)(j) = true
            return 1 + dfs(visited, m, n, k, i+1, j) + dfs(visited, m, n, k, i, j+1)
        }
    
        def sum(x: Int): Int = {
            var sum = 0
            var tempX = x
            while (tempX > 0) {
                sum += tempX % 10
                tempX /= 10
            }
            return sum
        }
    }
    
    
    //BFS
    import scala.collection.mutable.Queue
    import util.control.Breaks._
    object Solution {
        def movingCount(m: Int, n: Int, k: Int): Int = {
            val queue = Queue[(Int, Int)]()
            val visited = Array.fill(m, n)(false)
            var res = 0
    
            queue.enqueue((0, 0))
            while(queue.size > 0) {
                val a = queue.dequeue
                val x = a._1
                val y = a._2
                
                breakable{
                    if (x < 0 || x >= m || y < 0 || y >= n || visited(x)(y) == true || sum(x) + sum(y) > k) break()
                    visited(x)(y) = true
                    res += 1
                    queue.enqueue((x+1, y))
                    queue.enqueue((x, y+1))
                }
            }
            return res
        }
    
        def sum(i: Int): Int = {
            var sum = 0
            var tempI = i
            while (tempI > 0) {
                sum += tempI%10
                tempI /= 10
            }
            return sum
        }
    }
    
    //DFS
    func movingCount(m int, n int, k int) int {
        //visited := [m][n]bool{}
        visited := make([][]bool, m)
        for i := 0; i < m; i++ {
            visited[i] = make([]bool, n)
        }
    
        return dfs(visited, m, n, k, 0, 0)
    }
    
    func dfs(visited [][]bool, m, n, k, i, j int) int {
        if i < 0 || i >= m || j < 0 || j >= n || visited[i][j] == true || sum(i) + sum(j) > k { return 0}
        visited[i][j] = true
        return 1 + dfs(visited, m, n, k, i+1, j) + dfs(visited, m, n, k, i, j+1)
    } 
    
    func sum(x int) int {
        res := 0
        for (x > 0) {
            res += x % 10
            x /= 10
        }
        return res
    }
    
    //BFS
    import "container/list"
    func movingCount(m int, n int, k int) int {
        res := 0
        visited := make([][]bool, m)
        for i := 0; i < m; i++ {
            visited[i] = make([]bool, n)
        }
    
        queue := list.New()
        queue.PushBack(0)
        queue.PushBack(0)
    
        for (queue.Len() > 0) {
            temp := queue.Front()
            x := temp.Value.(int)
            queue.Remove(temp)
            temp = queue.Front()
            y := temp.Value.(int)
            queue.Remove(temp)
    
            if x < 0 || x >= m || y < 0 || y >= n || visited[x][y] == true ||sum(x) + sum(y) > k {continue}
            visited[x][y] = true
            res += 1
            queue.PushBack(x+1)
            queue.PushBack(y)
            queue.PushBack(x)
            queue.PushBack(y+1) 
        }
    
        return res
    }
    
    func sum(x int) int {
        res := 0 
        for (x > 0) {
            res += x%10
            x /= 10
        }
        return res
    }
    
  • 相关阅读:
    TCP/IP报文 三次握手 四次挥手
    socket 编程
    出现线程死锁的几种情况
    类模板的写法
    【HTTP】boundary 中一个 = 导致HTTP上传文件失败
    【时间戳】 年月日 转换为时间戳
    【CSV文件】CSV文件内容读取
    std::string 的方法c_str() 和 data() 有什么区别
    [转载] C++ STL中判断list为空,size()==0和empty()有什么区别
    【SQL】glob 和 like 的区别
  • 原文地址:https://www.cnblogs.com/ganshuoos/p/14151050.html
Copyright © 2011-2022 走看看