第九届蓝桥杯省赛——3字母阵列(二维数组)

这篇具有很好参考价值的文章主要介绍了第九届蓝桥杯省赛——3字母阵列(二维数组)。希望对大家有所帮助。如果存在错误或未考虑完全的地方,请大家不吝赐教,您也可以点击"举报违法"按钮提交疑问。

题目:

仔细寻找,会发现:在下面的8x8的方阵中,隐藏着字母序列:“LANQIAO”。

SLANQIAO

ZOEXCCGB

MOAYWKHI

BCCIPLJQ

SLANQIAO

RSFWFNYA

XIFZVWAL

COAIQNAL

我们约定: 序列可以水平,垂直,或者是斜向;

并且走向不限(实际上就是有一共8种方向)。

上图中一共有4个满足要求的串。

下面有一个更大的(100x100)的字母方阵。

你能算出其中隐藏了多少个“LANQIAO”吗?

FOAIQNALWIKEGNICJWAOSXDHTHZPOLGYELORAUHOHCZIERPTOOJUITQJCFNIYYQHSBEABBQZPNGYQTCLSKZFCYWDGOAIADKLSNGJ
GSOZTQKCCSDWGUWAUOZKNILGVNLMCLXQVBJENIHIVLRPVVXXFTHQUXUAVZZOFFJHYLMGTLANQIAOQQILCDCJERJASNCTLYGRMHGF
TSDFYTLVIBHKLJVVJUDMKGJZGNNSTPVLCKTOFMUEUFSVQIAUVHNVFTGBDDARPKYNNCUOYUAZXQJNOEEYKLFRMOEMHUKJTPETHLES
FKVINSLWEVGAGMKVFVIUBMYOIAFHLVNKNTYKTZWVXQWRWIGPENFXYDTKRVPKRTHMGHVYOCLDCKMEKRLGEKBYUCLOLYPAKPFSOREH
KWPUOLOVMOFBIXYACWRDYBINTMPASPCEOKHXQIGBQQMCEOIVULIEOPFSCSIHENAJCVDPJDOIWIIULFDNOZOFVAMCABVGKAKCOZMG
XWMYRTAFGFOCNHLBGNGOXPJSTWLZUNNAGIRETGXFWAQSSJPFTQAXMTQWMZWYVEPQERKSWTSCHSQOOBGXAQTBCHOEGBDVKGWJIFTG
ZWWJEIISPLMXIMGHOOGDRZFTGNDDWDWMNUFWJYJGULPHNUFSAQNNIUVAAFZIAZKFXXNWCEABGJAUMGYEIEFVQXVHHHEDYUITRCQB
XZHDPZQTOBECJVBZLACVXACZEDYOGVAVQRNWEOWGRAQYUEUESTEDQTYJUTEFOOITSHDDZHONJGBRCWNEQLZUTBNQIADKNFIOMWZR
EBFKCVNLURZSNPOLTISRPDTNUMCDGKTYRGIOVEPTUTSBAWQKWWEUWIWHAANUZUADGZEATZOQICWFUJTWNZDBKLQNELWVTBNDLNFH
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LVHAMSNXJOCVRPVGJUWXFVOCUCLCZDXRPBBDRLRAVVNLOZWOHWMXYSNMXAKJYWYGILNGUJGIPKAUDVANZLFWKUWWUSQYBRCBVDIJ
QCXPLOTPPGXCUZOUSSTXHVMLHVMJTUSSOPLRKEBQSGWNGVHKANVZWYQHSHLIPWSYCPKTUKPMWPLVFLLAHXZQANFXHFNYHIQVIOYN
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OQETOJHCZXGTUKIWGMEVVMXCURISUOFQSAWZWDMZWVYHZMPEIMWKJDGERODVVUXYRTYLCRGYQQOIOFZSSZRAIESWBQOAIQNALJNR
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分析:

要我们求八个方位连着的字符串,我们需要把八个方位的相对增喊值数组写出来,然后二维数组里面只要有等于L的,那么直接进行八个方位判断,每次x,y的坐标根据dx和dy发生变化,只要有一个相等那么num加一直到num等于6时,count加一。八个方位判断完毕,再进行下一个zu[i][j]==ch[0]的

check里num从1开始的原因是因为进入函数之前就进行一次if判断了,所以直接从第二个值开始文章来源地址https://www.toymoban.com/news/detail-608735.html

步骤:

第九届蓝桥杯省赛——3字母阵列(二维数组),蓝桥杯刷题,蓝桥杯,算法,职场和发展,Powered by 金山文档
package 第九届省赛;

import java.util.Scanner;

public class 字母阵列 {
    static int n = 100;
    static int[] dx = { 0, 1, 1, 1, 0, -1, -1, -1 };
    static int[] dy = { 1, 1, 0, -1, -1, -1, 0, 1 };
    static char zu[][] = new char[n][n];//用于储存扫描器给的字母阵列
    static char ch[] = { 'L', 'A', 'N', 'Q', 'I', 'A', 'O' };//需要找的字符串
    static int count = 0;//计数
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        for (int i = 0; i < zu.length; i++) {//通过扫描器给二维数组赋值
            zu[i] = sc.nextLine().toCharArray();
        }
        for (int i = 0; i < zu.length; i++) {
            for (int j = 0; j < zu[i].length; j++) {//遍历所有二维数组的值,只要等于ch的第一个那么进行八个方位的判断
                if (zu[i][j] == ch[0]) {
                    check(i, j);
                }
            }
        }
        System.out.println(count);
        sc.close();
    }
    private static void check(int i, int j) {
        for (int j2 = 0; j2 < 8; j2++) {
            int x=i+dx[j2];
            int y=j+dy[j2];
            int num = 1;
            while (x >= 0 && y >= 0 && x < n && y < n && zu[x][y] == ch[num]) {
                if (num == 6) {
                    count++;
                    break;
                }
                num++;
                x += dx[j2];
                y += dy[j2];
            }
        }

    }

}

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