首先,这是额外信用作业的一部分,所以请不要给我答案 . 请帮助我了解我可能遇到的问题 . 它是一个Tic-Tac-Toe发生器,游戏通过递归来确定基于玩家的最佳移动 . (教授用白色'W'和黑色'B'代替X和O)
我的主递归方法根据TTT板上的输入位置返回状态分数; 1如果白色将强制从该位置获胜,如果是平局则为0,如果黑色将强制从该位置获胜,则为-1:
public int stateScore(boolean whiteMove, int[] BestMove) {
return stateScore(whiteMove,BestMove,TTTBoard);
}
它调用我的底层私有递归方法:
private int stateScore(boolean whiteMove, int[] BestMove,char[][] TestBoard) {
char [][] newTestBoard = new char [3][3];
for(int rowVal = 0; rowVal < 3; rowVal++){
for(int colVal = 0; colVal < 3; colVal++){
newTestBoard[rowVal][colVal] = TestBoard[rowVal][colVal];
}
}
int [] bestMove = new int [2];
for(int rowVal = 0; rowVal < 3; rowVal++){
for(int colVal = 0; colVal < 3; colVal++){
if(isFull(newTestBoard) == true){
return 0;
}
else if(newTestBoard[rowVal][colVal] == '-'){
bestMove[0] = rowVal;
bestMove[1] = colVal;
//if boolean is white
if(whiteMove == true){
newTestBoard = testEntry(rowVal,colVal,'W',newTestBoard);
if(threeInRow(newTestBoard) == 1){
return 1;
}
else if(threeInRow(newTestBoard) == 0 && isFull(newTestBoard) == true){
return 0;
}
else if(threeInRow(newTestBoard) == -1 && isFull(newTestBoard) == true){
return -1;
}
else{
return stateScore(!whiteMove,bestMove,newTestBoard);
}
}
//if boolean is black
else{
newTestBoard = testEntry(rowVal,colVal,'B',newTestBoard);
if(threeInRow(newTestBoard) == -1){
return -1;
}
else if(threeInRow(newTestBoard) == 0 && isFull(newTestBoard) == true){
return 0;
}
else if(threeInRow(newTestBoard) == 1 && isFull(newTestBoard) == true){
return 1;
}
else{
return stateScore(!whiteMove,bestMove);
}
}
}
}
}
return 0;
}
whiteMove的布尔值如果是白色移动则为true,如果是黑色则为false . 函数中的辅助方法包括threeInRow:
public int threeInRow(char[][] TTTBoard){
boolean whiteIs = false;
boolean blackIs = false;
//Horizontal?
char [] colChar = new char [3];
for(int rowVal = 0; rowVal < 3; rowVal ++){
for(int colVal = 0; colVal < 3; colVal++){
colChar[colVal] = TTTBoard[rowVal][colVal];
}
if(colChar[0] == colChar[1] && colChar[1] == colChar[2]){
if(colChar[0] == 'W'){
whiteIs = true;
}
if(colChar[0] == 'B'){
blackIs = true;
}
}
}
//Vertical?
char [] rowChar = new char [3];
for(int colVal = 0; colVal < 3; colVal ++){
for(int rowVal = 0; rowVal < 3; rowVal++){
rowChar[colVal] = TTTBoard[rowVal][colVal];
}
if(rowChar[0] == rowChar[1] && rowChar[1] == rowChar[2]){
if(rowChar[0] == 'W'){
whiteIs = true;
}
else if(rowChar[0] == 'B'){
blackIs = true;
}
}
}
//Diagonal
//topLeft to bottomRight
if(TTTBoard[0][0] == TTTBoard[1][1] && TTTBoard[1][1] == TTTBoard[2][2]){
if(TTTBoard[0][0] == 'W'){
whiteIs = true;
}
else if(TTTBoard[0][0] == 'B'){
blackIs = true;
}
}
//topRight to bottomLeft
if(TTTBoard[0][2] == TTTBoard[1][1] && TTTBoard[1][1] == TTTBoard [2][0]){
if(TTTBoard[1][1] == 'W'){
whiteIs = true;
}
else if(TTTBoard[1][1] == 'B'){
blackIs = true;
}
}
//Return Vals
if(whiteIs == true && blackIs == true){
return 0;
}
else if(blackIs == true && whiteIs == false){
return -1;
}
else if(blackIs == false && whiteIs == true){
return 1;
}
else if(blackIs == false && whiteIs == false){
return 0;
}
else{
return 0;
}
}
和testEntry:
public char[][] testEntry(int row,int col,char newChar, char[][] TestBoard){
char [][] returnBoard = new char[3][3];
for(int rowVal = 0; rowVal < 3; rowVal++){
for(int colVal = 0; colVal < 3; colVal++){
returnBoard[rowVal][colVal] = TestBoard[rowVal][colVal];
}
}
returnBoard[row][col] = newChar;
return returnBoard;
}
我没有使用带递归的for循环,我是不是搞乱了 . 另外,我说 type [] name = name
(相同类型)不起作用是对的,对吧?这就是为什么我在那种情况下做了for循环 .
2 回答
在你的黑色分支中,你的回归是错误的 .
你回来了
这会重新启动递归 . 你想回来
提示:
使用UpperCase for Classes,lowerCamelCase用于变量 .
如果你在if分支中返回,那么你不需要别的 .
代替:
最好写:
保持较低的嵌套,使代码更容易遵循 .
为什么不把它移到if(whiteMove)之外
发布堆栈跟踪,但我敢打赌,当你递归调用
stateScore
时,你会得到无限递归 .