我为所有方向创建了一个单词解算器。它可以水平，垂直和反向找到单词。但是，我遇到了问题，无法向所有方向发展。所以在“
H E i l
x L p q
c L O m
任何人都可以指出我该怎么做？这是我的算法搜索单词（在C ++中）：
/*
* For loops that search each row, each column in all 8 possible directions.
*/
void Scramble::solve() {
cout << "Output:" << endl;
for (int row = 0; row < getRows(); row++) {
for (int col = 0; col < getCols(); col++)
for (int rowDir = -1; rowDir <= 1; rowDir++)
for (int colDir = -1; colDir <=1; colDir++)
if (rowDir != 0 || colDir != 0)
findWords(row, col, rowDir, colDir);
}
}
/*
* Finds the matches in a given direction. Also calls verifyWord() to verify that the
* current sequence of letters could possibly form a word. If not, search stops.
*/
void Scramble::findWords(int startingRow, int startingCol, int rowDir, int colDir) {
int searchResult;
string sequence = "";
sequence = sequence + wordsArr[startingRow][startingCol];
for (int i = startingRow + rowDir, j = startingCol + colDir; i >= 0 && j >= 0
&& i < getRows() && j < getCols(); i = i + rowDir, j = j + colDir) {
sequence = sequence + wordsArr[i][j];
if (sequence.length() >= 3) {
searchResult = verifyWord(words, sequence);
if ((unsigned int)searchResult == words.size())
break;
if (words[searchResult].rfind(sequence) > words[searchResult].length())
break;
if (words[searchResult] == (sequence))
cout << sequence << endl;
}
}
}
/*
* Performs the verifyWord search method.
* Searches the word to make sure that so far, there is possibly that the current sequence
* of letter could form a word. That is to avoid continuing to search for a word
* when the first sequence of characters do not construct a valid word in the dictionary.
*
* For example, if we have 'xzt', when this search is done it prevents the search
* to continue since no word in the dictionary starts with 'xzt'
*/
int Scramble::verifyWord(vector<string> words, string str) {
int low = 0;
int mid = 0;
int high = words.size();
while (low < high) {
mid = (low + high) / 2;
if (str > words[mid]) {
low = mid + 1;
}
else if (str < words[mid]) {
high = mid - 1;
}
else
return mid;
}
}
答案 0 :(得分：2)
这是一个有趣的思考方式：找到这个词类似于解决迷宫。 'start'和'end'对应于您正在寻找的单词的开头和结尾，'dead end'对应于路径和单词之间的不匹配，'success'表示沿着您的路径的字符串是一场比赛。
这里的好消息是迷宫解决算法有很多资源。我熟悉并且不太难实现的一种特殊算法是recursion with backtracking。
显然，必须进行一些更改才能使其适用于您的问题。例如，你不知道起点在哪里，但幸运的是它并不重要。您可以检查每个可能的起始位置，由于不匹配，其中许多将在第一步被丢弃。
答案 1 :(得分：0)
只需将其视为图形，其中每个字母都连接到所有相邻字母，并从每个字母开始执行深度/广度优先搜索，仅接受那些字母等于您要查找的下一个字母的节点
答案 2 :(得分：0)
1）目前，您的solve()函数从每个点开始在一条直线中查找单词：这是您想要的吗？我只是问，因为'你好'不会在你的样本矩阵中显示为直线：
H E i l
x L p q
c L O m
如果你确实只想要直线字样，那么很好（这就是我总是理解these puzzles无论如何都要工作的方式），但实际上你想找到单词在蛇方式方式中，然后像Zilchonum和BlueRaja这样的递归搜索表明这将是一个不错的选择。请注意，不要以你已经使用过的字母循环回来。
2）在任何一种情况下，您的verifyWord()函数也存在一些问题：在退出while (low < high)循环的情况下，至少需要返回一些值。
即便如此，它仍然不会完全按照您的意愿行事：例如，说出你的字典
包含{"ant", "bat" "hello", "yak", "zoo"}，并且您使用verifyWord()调用str="hel"，您希望返回值2，但此时它会执行此操作：
step low mid high
0 0 0 5 // initialise
1 0 2 5 // set mid = (0+5)/2 = 2... words[2] == "hello"
2 0 2 1 // "hel" < "hello" so set high = mid - 1
3 0 0 1 // set mid = (0+1)/2 = 0... words[0] == "ant"
4 1 0 1 // "hel" > "ant" so set low = mid + 1
5 // now (low<high) is false, so we exit the loop with mid==0
与其将“hel”与“hello”进行比较，或许最好将字典中的单词截断为与str相同的长度：即将str与word[mid].substr(0,str.length())进行比较？
答案 3 :(得分：0)
这是我写的一个简单的单词侦探程序---＆gt;
#include<iostream>
using namespace std;
int main()
{
int a, b, i, j, l, t, n, f, g, k;
cout<<"Enter the number of rows and columns: "<<endl;
cin>>a>>b; //Inputs the number of rows and columns
char mat[100][100], s[100];
cout<<"Enter the matrix: "<<endl;
for (i = 0; i < a; i++) for (j = 0; j < b; j++) cin>>mat[i][j]; //Inputs the matrix
cout<<"Enter the number of words: "<<endl;
cin>>t; //Inputs the number of words to be found
while (t--)
{
cout<<"Enter the length of the word: "<<endl;
cin>>n; //Inputs the length of the word
cout<<"Enter the word: "<<endl;
for (i = 0; i < n; i++) cin>>s[i]; //Inputs the word to be found
for (i = 0; i < a; i++) //Loop to transverse along i'th row
{
for (j = 0; j < b; j++) //Loop to transverse along j'th column
{
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, g++); //Loop to find the word if it is horizontally right
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" right"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, g--); //Loop to find the word if it is horizontally left
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" left"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f++); //Loop to find the word if it is vertically down
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f--); //Loop to find the word if it is vertically up
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f++, g++); //Loop to find the word if it is down right
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down right"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f--, g--); //Loop to find the word if it is up left
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up left"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f++, g--); //Loop to find the word if it is down left
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down left"<<endl;
goto A;
}
f = i;
g = j;
for (k = 0; s[k] == mat[f][g] && k < n; k++, f--, g++); //Loop to find the word if it is up right
if (k == n)
{
cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up right"<<endl;
goto A;
}
}
}
A:; //If the word has been found the program should reach this point to start the search for the next word
}
return 0;
}
在我的程序中，它首先检查单词的第一个字母，然后检查后续字母。如果找到该单词，则会打印单词的起始坐标和找到单词的方向。