Google Kick Start 2019 C轮 第一题 Wiggle Walk 题解

Google Kick Start 2019 C轮 第一题 Wiggle Walk 题解

题目地址：https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050ff2/0000000000150aac

四个解法：

1. 暴力模拟
2. 使用HashMap优化，理论时间复杂度最小（好像也是并查集）
3. (推荐)使用BitSet，实际占用空间特别小，仅仅是 2mn 个比特大小
4. 使用HashMap实现的并查集方法，在东南西北4个方向上用并查集处理

解法1：暴力模拟

下面的代码是我写的暴力模拟办法，只能过样例，提交上去会 Runtime Error.

import java.util.*;
import java.io.*;
public class Solution {
    static String input = "3" + "\n"
        + "5 3 6 2 3" + "\n"
        + "EEWNS" + "\n"
        + "4 3 3 1 1" + "\n"
        + "SESE" + "\n"
        + "11 5 8 3 4" + "\n"
        + "NEESSWWNESE" + "\n";

    public static void main(String[] args) {
        //Scanner sc = new Scanner(System.in);
        Scanner sc = new Scanner(new StringReader(input));
        int T = sc.nextInt();
        int[][] D = new int[128][2];
        D['E'] = new int[]{0, 1};
        D['W'] = new int[]{0, -1};
        D['N'] = new int[]{-1,0};
        D['S'] = new int[]{1,0};
        for (int i = 0; i < T; i++) {
            int N = sc.nextInt();
            int R = sc.nextInt();
            int C = sc.nextInt();
            int Sr = sc.nextInt();
            int Sc = sc.nextInt();
            sc.nextLine();
            String seq = sc.nextLine();
            boolean[][] map = new boolean[R + 1][C + 1];
            map[Sr][Sc] = true;

            for (int j = 0; j < N; j++) {
                do {
                    Sr += D[seq.charAt(j)][0];
                    Sc += D[seq.charAt(j)][1];
                } while (map[Sr][Sc] == true);
                map[Sr][Sc] = true;
            }
            System.out.println("Case #" + (i + 1) + ": " + Sr + " " + Sc);
        }
    }
}

解法2：HashMap优化

下面是使用HashMap能过的版本，原答案来自StackExchange上的Python解法.

注意静态内部类 Point，我重写了对象的equals()方法和hashCode()方法，以便 HashMap 能正确查找对象。hashCode()方法用于计算哈希值，不重写的会采用对象内存地址作为哈希值，这是我们不希望看到的。equals()方法用于解决哈希冲突后的对象实际内容比对。

import java.util.*;
import java.io.*;
public class Solution {

    static class Point {
        int x;
        int y;
        Point() {}
        Point(int xx, int yy) { x = xx; y = yy; }
        @Override
        public boolean equals(Object obj) {
            Point that = (Point) obj;
            return this.x == that.x && this.y == that.y;
        }
        @Override
        public int hashCode() {
            return x * 50000  + y;
        }
    }
    static HashMap<Point, Point>[] neighbors;

    public static void init() {
        neighbors = new HashMap[128];
        neighbors['W'] = new HashMap<Point, Point>();
        neighbors['E'] = new HashMap<Point, Point>();
        neighbors['S'] = new HashMap<Point, Point>();
        neighbors['N'] = new HashMap<Point, Point>();
    }

    public static Point getNeighbor(Point cur, char direction) {
        if (neighbors[direction].containsKey(cur)) {
            return neighbors[direction].get(cur);
        }
        switch(direction) {
            case 'W': return new Point(cur.x - 1, cur.y);
            case 'E': return new Point(cur.x + 1, cur.y);
            case 'N': return new Point(cur.x, cur.y - 1);
            case 'S': return new Point(cur.x, cur.y + 1);
            default: return null;
        }
    }

    public static void linkNeighbors(Point cur) {
        Point west = getNeighbor(cur, 'W');
        Point east = getNeighbor(cur, 'E');
        Point north = getNeighbor(cur, 'N');
        Point south = getNeighbor(cur, 'S');
        neighbors['W'].put(east, west);
        neighbors['E'].put(west, east);
        neighbors['N'].put(south, north);
        neighbors['S'].put(north, south);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int N = sc.nextInt();
            int R = sc.nextInt();
            int C = sc.nextInt();
            int Sr = sc.nextInt();
            int Sc = sc.nextInt();
            sc.nextLine();  // skip \n at end of previous line
            String seq = sc.nextLine();

            init();
            Point cur = new Point(Sc, Sr);
            for (int j = 0; j < N; j++) {
                linkNeighbors(cur);
                cur = getNeighbor(cur, seq.charAt(j));
            }

            System.out.println("Case #" + (i + 1) + ": " + cur.y + " " + cur.x);
        }
    }
}

解法3: BitSet

思路来源： http://codeforces.com/blog/entry/67224?#comment-513594

核心就是利用 BitSet 提供的 previousSetBit() 和 nextSetBit() 方法，虽然这两个方法理论上是 O(n) 时间复杂度，但由于是位操作，且Java肯定做了某些特殊优化，所有不仅占用内存特别小，运行速度也快。

import java.util.*;
import java.io.*;
public class Solution {

    static BitSet[] rowBitsets;   // m horizontal bitset
    static BitSet[] colBitsets;   // n vertical bitsets

    public static void init(int rows, int cols) {
        // initilize m + n bitsets
        rowBitsets = new BitSet[rows + 1];
        colBitsets = new BitSet[cols + 1];
        for (int i = 1; i <= rows; i++) rowBitsets[i] = new BitSet(cols + 1);
        for (int i = 1; i <= cols; i++) colBitsets[i] = new BitSet(rows + 1);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int N = sc.nextInt();
            int R = sc.nextInt();
            int C = sc.nextInt();
            int Sr = sc.nextInt();
            int Sc = sc.nextInt();
            sc.nextLine();  // skip \n at end of previous line
            String seq = sc.nextLine();

            init(R, C);
            for (int j = 0; j < N; j++) {
                rowBitsets[Sr].set(Sc);
                colBitsets[Sc].set(Sr);

                switch(seq.charAt(j)) {
                    case 'W': Sc = rowBitsets[Sr].previousClearBit(Sc); break;
                    case 'E': Sc = rowBitsets[Sr].nextClearBit(Sc); break;
                    case 'N': Sr = colBitsets[Sc].previousClearBit(Sr); break;
                    case 'S': Sr = colBitsets[Sc].nextClearBit(Sr); break;
                    default: break;
                }
            }

            System.out.println("Case #" + (i + 1) + ": " + Sr + " " + Sc);
        }
    }
}

解法4：使用并查集

import java.util.*;
import java.io.*;
public class Solution {

    static class Point {
        int x, y;
        Point(int xx, int yy) { x = xx; y = yy; }
        @Override
        public boolean equals(Object obj) {
            Point that = (Point) obj;
            return this.x == that.x && this.y == that.y;
        }
        @Override
        public int hashCode() {
            return x * 50001 + y;
        }
    }

    public static Point findNeighbor(HashMap<Point, Point> map, Point cur) {
        // 向某个方向查找
        if (!map.containsKey(cur) || map.get(cur).equals(cur)) {
            // 当前结点cur在目标方向上没有已知的邻居，返回当前结点
            return cur;
        } else {
            // 当前结点cur在目标方向上有个邻居，在目标方向上查找邻居的邻居，直到找到最远的邻居
            // 并将当前结点在目标方向上的邻居更新为最远邻居，返回最远邻居
            Point res = findNeighbor(map, map.get(cur));
            map.put(cur, res);
            return res;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int N = sc.nextInt();
            int R = sc.nextInt();
            int C = sc.nextInt();
            int Sr = sc.nextInt();
            int Sc = sc.nextInt();
            sc.nextLine();  // skip \n at end of previous line
            String seq = sc.nextLine();

            HashMap<Point, Point> mapN = new HashMap<>();
            HashMap<Point, Point> mapS = new HashMap<>();
            HashMap<Point, Point> mapW = new HashMap<>();
            HashMap<Point, Point> mapE = new HashMap<>();
            Point cur = new Point(Sr, Sc);
            for (int j = 0; j < N; j++) {
                // 实在是搞不懂这句
                mapN.put(new Point(cur.x + 1, cur.y), cur);
                mapS.put(new Point(cur.x - 1, cur.y), cur);
                mapW.put(new Point(cur.x, cur.y + 1), cur);
                mapE.put(new Point(cur.x, cur.y - 1), cur);
                // 更新4个点
                switch(seq.charAt(j)) {
                    case 'N':
                        Point t1 = findNeighbor(mapN, cur); // 找到当前结点的最远邻居
                        cur = new Point(t1.x - 1, t1.y);
                        break;
                    case 'S':
                        Point t2 = findNeighbor(mapS, cur); // 找到当前结点的最远邻居
                        cur = new Point(t2.x + 1, t2.y);
                        break;
                    case 'E':
                        Point t3 = findNeighbor(mapE, cur); // 找到当前结点的最远邻居
                        cur = new Point(t3.x, t3.y + 1);
                        break;
                    case 'W':
                        Point t4 = findNeighbor(mapW, cur); // 找到当前结点的最远邻居
                        cur = new Point(t4.x, t4.y - 1);
                        break;
                    default: break;
                }
                System.out.println(seq.charAt(j) + " " + cur.x + " " + cur.y);
            }

            System.out.println("Case #" + (i + 1) + ": " + cur.x + " " + cur.y);
        }
    }
}