• 剑指offer: JZ13 机器人的运动范围


    DFS、BFS解决JZ13机器人的运动范围

    描述

    地上有一个 rows 行和 cols 列的方格。坐标从 [0,0] 到 [rows-1,cols-1] 。一个机器人从坐标 [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,但是不能进入行坐标和列坐标的数位之和大于 threshold 的格子。 例如,当 threshold 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? 输入:1,2,3返回值:3

    分析

    通过打印这个地图的坐标(0代表可行走1代表不可行走)可以看出,通过回溯dfs的方法就可以计算出所能走的路径,以及不断计算节点周围能走动的节点得出答案bfs,本人采用了两种方案分别实现了本题目

    代码

        /**
         * JZ13 机器人的运动范围
         * @param threshold
         * @param rows
         * @param cols
         * @return
         */
        public int movingCount(int threshold, int rows, int cols) {
            int maps[][] = new int[rows][cols];
            //通过输入的值进行地图的二维数组初始化
            for (int i = 0;i<maps.length;i++) {
                for (int j = 0;j<maps[0].length;j++) {
                    if ((i/100+i/10+i % 10) + (j/100+j/10+j%10)>threshold)
                        maps[i][j] = 1;
                }
            }
            //打印地图供理解
            for (int i = 0;i<maps.length;i++) {
                for (int j = 0;j< maps[i].length;j++) {
                    System.out.print(maps[i][j]);
                }
                System.out.println();
            }
            //初始化地图的每个节点
            MoveMapNode nodesMap[][] = new MoveMapNode[rows][cols];
            for (int i = 0;i<maps.length;i++) {
                for (int j =0;j<maps[i].length;j++) {
                    MoveMapNode node = new MoveMapNode(maps[i][j],i,j,maps);
                    nodesMap[i][j] = node;
                }
            }
    
            return movingCountBFS(nodesMap);
        }
    
        /**
         * bfs实现
         * @param nodesMap
         * @return
         */
        public int movingCountBFS(MoveMapNode nodesMap[][]) {
            Queue<MoveMapNode> nodesQueue = new LinkedList<>();
            nodesQueue.offer(nodesMap[0][0]);
            int count = 1;
            nodesMap[0][0].canVisit = false;
            while (nodesQueue.size()>0) {
                //上
                if (nodesQueue.peek().x > 0 && nodesMap[nodesQueue.peek().x - 1][nodesQueue.peek().y].canVisit == true)  {
                    nodesMap[nodesQueue.peek().x - 1][nodesQueue.peek().y].canVisit = false;
                    nodesQueue.offer(nodesMap[nodesQueue.peek().x - 1][nodesQueue.peek().y]);
                    count ++;
                }
                //下
                if (nodesQueue.peek().x < nodesMap.length - 1 && nodesMap[nodesQueue.peek().x + 1][nodesQueue.peek().y].canVisit == true) {
                    nodesMap[nodesQueue.peek().x + 1][nodesQueue.peek().y].canVisit = false;
                    nodesQueue.offer(nodesMap[nodesQueue.peek().x + 1][nodesQueue.peek().y]);
                    count ++;
                }
                //左
                if (nodesQueue.peek().y > 0 && nodesMap[nodesQueue.peek().x][nodesQueue.peek().y - 1].canVisit == true) {
                    nodesMap[nodesQueue.peek().x][nodesQueue.peek().y - 1].canVisit = false;
                    nodesQueue.offer(nodesMap[nodesQueue.peek().x][nodesQueue.peek().y - 1]);
                    count ++;
                }
                //右
                if (nodesQueue.peek().y < nodesMap[0].length - 1 && nodesMap[nodesQueue.peek().x][nodesQueue.peek().y + 1].canVisit == true) {
                    nodesMap[nodesQueue.peek().x][nodesQueue.peek().y + 1].canVisit = false;
                    nodesQueue.offer(nodesMap[nodesQueue.peek().x][nodesQueue.peek().y + 1]);
                    count ++;
                }
                nodesQueue.poll();
            }
    
            return count;
        }
        /**
         * dfs实现
         * @param nodesMap
         * @return
         */
        public int movingCountDFS(MoveMapNode nodesMap[][]) {
    
            Stack<MoveMapNode> stack = new Stack<>();
            stack.push(nodesMap[0][0]);
            int count = 1;
            while (stack.size() > 0) {
                if (stack.peek().x - 1 > 0 && nodesMap[stack.peek().x - 1][stack.peek().y].canVisit == true) {
                    //向上判断
                    stack.push(nodesMap[stack.peek().x - 1][stack.peek().y]);
                    stack.peek().canVisit = false;
                    count ++;
                }else if (stack.peek().x + 1 <nodesMap.length && nodesMap[stack.peek().x + 1][stack.peek().y].canVisit == true) {
                    //向下判断
                    stack.push(nodesMap[stack.peek().x + 1][stack.peek().y]);
                    stack.peek().canVisit = false;
                    count ++;
                }else if (stack.peek().y - 1 > 0 && nodesMap[stack.peek().x][stack.peek().y - 1].canVisit == true) {
                    //向左判断
                    stack.push(nodesMap[stack.peek().x][stack.peek().y - 1]);
                    stack.peek().canVisit = false;
                    count ++;
                }else if (stack.peek().y + 1 <  nodesMap[0].length && nodesMap[stack.peek().x][stack.peek().y + 1].canVisit == true) {
                    //向右判断
                    stack.push(nodesMap[stack.peek().x][stack.peek().y + 1]);
                    stack.peek().canVisit = false;
                    count ++;
                }else {
                    stack.pop();
                }
            }
            return count;
        }
    
        /**
         * 每个节点的实体类
         */
        class MoveMapNode{
            int number;
            int x;
            int y;
    
            boolean canVisit = true;
    
            //如果当前节点为1就代表不能走动,置canVisit为false
            MoveMapNode(int number,int x,int y,int[][]map) {
                this.number = number;
                this.x = x;
                this.y = y;
    
                if (number == 1) {
                    canVisit = false;
                }
            }
        }
    
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  • 原文地址:https://www.cnblogs.com/bearcanlight/p/15505822.html
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