我从用户那里获取单词列表作为输入并将其输入存储在String []数组中。然后，我创建一个名为letterGrid的char [] []数组，其中填充了随机字母和用户提供的单词。然后，当在控制台屏幕上显示letterGrid时，用户必须输入他们想要查找的单词。然后，程序将检查输入的单词是水平，垂直还是对角线出现，并打印出红色，橙色，黄色，绿色，蓝色，紫色，彩虹色。输出网格：
如您所见，当我输入黄色（长度为6个字母）时，程序将输出单词的位置，但是会出现错误。
编辑代码
以下是@Igor Khvostenkov要求的其余代码：
private String word; // This variable will be the user`s input when they chose to search for a word they have entered
private int rowLocation; // This variable will represent the row number in which the word is at
private int colLocation; // This variable will represent the column number in which the word is at
// Create a method to compare the user`s word to the elements in the letter grid
public void compare (String word) {
for (int row = 0; row < letterGrid.length - 1; row++) {
for (int col = 0; col < letterGrid[row].length - 1; col++) {
if (letterGrid[row][col] == word.charAt(0)) {
rowLocation = row;
colLocation = col;
wordContains(); // Call on method to see if the word entered by the user appears horizontally, vertically, or diagonally
}//end of if
}//end of inner for loop
}//end of outer for loop
}//end of compare(word)
// Create a method that will check the direction of the user`s word input
public void wordContains() {
checkHorizontal(); // Checking if the word appears horizontally
checkVertical(); // Checking id the word appears vertically
checkDiagonal(); // Checking if the word appears diagonally
}//end of wordContains()
// Create a method to check if the user`s word appears HORIZONTALLY in the letter grid
public void checkHorizontal() {
for (int i = 1; i < (word.length()); i++) {
if (colLocation + i > letterGrid[0].length - 1) {
return;
} else if(letterGrid[rowLocation][colLocation + i] != word.charAt(i)) {
return;
}//end of if..else if
}//end of for loop
System.out.println(word + " found horizontally at row " + rowLocation + " and column " + colLocation); // Word found!!
System.out.println();
return;
}//end of checkHorizontal()
// Create a method to check if the user`s word appears VERTICALLY in the letter grid
public void checkVertical() {
for (int i = 1; i < (word.length()); i++) {
if (rowLocation + i > letterGrid.length - 1 && colLocation + i > letterGrid[0].length) {
return;
} else if (letterGrid[rowLocation + i][colLocation] != word.charAt(i)) {
return;
}//end of if..else if
}//end of for loop
System.out.println(word + " found vertically at row " + rowLocation + " and column " + colLocation); // Word found!!
System.out.println();
}//end of checkVertical()
// Create a method to check if the user`s word appears DIAGONALLY in the letter grid
public void checkDiagonal() {
for (int i = 1; i < (word.length()); i++) {
if (colLocation + i > letterGrid[0].length - 1 || rowLocation + i > letterGrid.length - 1) {
return;
} else if (letterGrid[rowLocation + i][colLocation + i] != word.charAt(i)) {
return;
}//end of if..else if
}//end of for loop
System.out.println(word + " found diagonally at row " + rowLocation + " and column " + colLocation); // Word found!!
System.out.println("");
}//end of checkDiagonal()我可以一直试图理解问题并寻找帮助我解决问题的方法。
最佳答案
代码中的问题是您在checkVertical()中的if条件中，行和列可能是第一个还是最后一个，但是您应该检查行或列。您的黄色示例失败，因为第一个代码在第一行和第二列中找到了Y，然后继续扫描，最后，它在最后一行中找到了Y，并检查了跳过的rowLocation + i > letterGrid.length - 1 && colLocation + i > letterGrid[0].length，然后调用了else将1加到该行，结果数组超出范围。这应该工作：
public void checkVertical() {
for (int i = 1; i < (word.length()); i++) {
if (rowLocation + i > letterGrid.length - 1 || colLocation + i > letterGrid[0].length) {
return;
} else if (letterGrid[rowLocation + i][colLocation] != word.charAt(i)) {
return;
}//end of if..else if
}//end of for loop
System.out.println(word + " found vertically at row " + rowLocation + " and column " + colLocation); // Word found!!
System.out.println();
}//end of checkVertical()