标题：字母阵列

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？

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GSOZTQKCCSDWGUWAUOZKNILGVNLMCLXQVBJENIHIVLRPVVXXFTHQUXUAVZZOFFJHYLMGTLANQIAOQQILCDCJERJASNCTLYGRMHGF
TSDFYTLVIBHKLJVVJUDMKGJZGNNSTPVLCKTOFMUEUFSVQIAUVHNVFTGBDDARPKYNNCUOYUAZXQJNOEEYKLFRMOEMHUKJTPETHLES
FKVINSLWEVGAGMKVFVIUBMYOIAFHLVNKNTYKTZWVXQWRWIGPENFXYDTKRVPKRTHMGHVYOCLDCKMEKRLGEKBYUCLOLYPAKPFSOREH
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HEYWHPLLPCUEOCBAOWGAYEJZQJHLVNMVQNSQQGGUBOIMDPFLOVSQGBLYAMBRYJDVOXOQINLJAVYALAKHPKOYNKGXIISSJNGKHYMS
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CURAGWXMLYBSIZLAXFONZZMQMRNNSRQKRHQGFGZUTLONAYRKSSOWAMKZBSGOOYQDPTBHGPBNQEDCZHRTOXREOFJEKJVIZXZBCJPN
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DDAOBVCPOVODVJFLSNPJXHWQBHILWZAHQQMTQASNADZLZNXJLJMFCOUWOZJCMVVTYCKTUBABWLCEBNYWAMOLNBQQYBRUJCQCZALE
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请提交一个整数，不要填写任何多余的内容。

解析一：

String lq = "LANQIAO";  //定义需要搜索匹配的字符串
char[][] arrChar = new char[100][100]; //定义100*100数组存放字符串
Scanner input = new Scanner(System.in);
//利用循环将字符串填充至数组，原始字符串正好以回车进行的每行分隔,
//所以只需要将原始字符串一次输入就等于100次输入
System.out.println("请将原始字符串复制粘贴到下方:");
for (int i = 0; i < 100; i++) {
    arrChar[i] = input.nextLine().toCharArray();
}
//定义二维数组保存8个移动方向的下标移动变化规律，从上开始，顺时针定义
int[][] arrMove = new int[][]{{0,-1},{1,-1},{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1}};
int result = 0;   //记录符合条件的次数
//循环所有字符
for (int i = 0; i < arrChar.length; i++) 
{
    for (int j = 0; j < arrChar[i].length; j++) 
    {
        if(arrChar[i][j] == lq.charAt(0))   //循环过程中发现字符为"L",则进行细节判断
        {
            //以此字符为中心，向8个方向去找复合条件的情况，循环arrMove的一维
            for (int k = 0;k < arrMove.length; k++) 
            {
                int rowIndex = i; //记录当前行号
                int colIndex = j;  //记录当前列号
                boolean flag = true;  //假设可以找到"LANQIAO"
                //向每个方向需要找6次,第一个L不需要查找
                for (int m = 1; m < lq.length(); m++) 
                {
                    rowIndex = rowIndex + arrMove[k][0];
                    colIndex = colIndex + arrMove[k][1];
                    //在移动过程中下标超出界限，一定找不到符合条件的
                    if(rowIndex < 0 || rowIndex >= 100 || colIndex < 0 || colIndex >= 100)
                    {
                        flag = false;  break;
                    }
                    //移动过程中将移动的出来的字符和"LANQIAO"中字符一个一个比较,不相等则退出
                    if(arrChar[rowIndex][colIndex] != lq.charAt(m))
                    {
                        flag = false; break;
                    }
                }
                //6次比对循环完成后flag值如果仍然为true，则证明符合条件
                if(flag == true)
                    result++;
            }
        }
    }
}
System.out.println(result); //打印最后结果

答案：41

备注：上述方案使用二维数组存储8个方向，除此之外，也可以定义一个类来存储方向，如下解析：

解析二：

class MyDirection
{
    int x;
    int y;
    public MyDirection(int x,int y)
    {
        this.x = x;
        this.y = y;
    }
}
public class Demo03 
{
    public static void main(String[] args)
    {
        String lq = "LANQIAO";  //定义需要搜索匹配的字符串
        char[][] arrChar = new char[100][100]; //定义100*100数组存放字符串
        Scanner input = new Scanner(System.in);
        //利用循环将字符串填充至数组，原始字符串正好以回车进行的每行分隔,
        //所以只需要将原始字符串一次输入就等于100次输入
        System.out.println("请将原始字符串复制粘贴到下方:");
        for (int i = 0; i < 100; i++) {
            arrChar[i] = input.nextLine().toCharArray();
        }
        //定义对象数组保存8个移动方向的下标移动变化规律，从上开始，顺时针定义
        MyDirection[] arrMove = new MyDirection[8];
        arrMove[0] = new MyDirection(0,-1);
        arrMove[1] = new MyDirection(1,-1);
        arrMove[2] = new MyDirection(1,0);
        arrMove[3] = new MyDirection(1,1);
        arrMove[4] = new MyDirection(0,1);
        arrMove[5] = new MyDirection(-1,1);
        arrMove[6] = new MyDirection(-1,0);
        arrMove[7] = new MyDirection(-1,-1);    
        int result = 0;   //记录符合条件的次数
        //循环所有字符
        for (int i = 0; i < arrChar.length; i++) 
        {
            for (int j = 0; j < arrChar[i].length; j++) 
            {
                if(arrChar[i][j] == lq.charAt(0))   //循环过程中发现字符为"L",则进行细节判断
                {
                    //以此字符为中心，向8个方向去找复合条件的情况，循环arrMove的一维
                    for (int k = 0;k < arrMove.length; k++) 
                    {
                        int rowIndex = i; //记录当前行号
                        int colIndex = j;  //记录当前列号
                        boolean flag = true;  //假设可以找到"LANQIAO"
                        //向每个方向需要找6次,第一个L不需要查找
                        for (int m = 1; m < lq.length(); m++) 
                        {
                            rowIndex = rowIndex + arrMove[k].x;
                            colIndex = colIndex + arrMove[k].y;
                            //在移动过程中下标超出界限，一定找不到符合条件的
                            if(rowIndex < 0 || rowIndex >= 100 || colIndex < 0 || colIndex >= 100)
                            {
                                flag = false;  break;
                            }
                            //移动过程中将移动的出来的字符和"LANQIAO"中字符一个一个比较,不相等则退出
                            if(arrChar[rowIndex][colIndex] != lq.charAt(m))
                            {
                                flag = false; break;
                            }
                        }
                        //6次比对循环完成后flag值如果仍然为true，则证明符合条件
                        if(flag == true)
                            result++;
                    }
                }
            }
        }
        System.out.println(result); //打印最后结果        
    }
}