java 数独 回溯 递归

Question

嘿，我写了这个程序来解决数独问题，但它只适用于数独矩阵的几个单元格，而其他单元格则返回 0 。你能明白这有什么问题吗？ 我是 Java 编码新手，无法编写简单的程序真的很痛苦。

public class sudoku {

static int sud[][] = new int[9][9];

public static void main(String args[]) {

    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            sud[i][j] = 0;
        }
    }
    solve(sud);
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            System.out.print(sud[i][j]);
        }
        System.out.print("\n");
    }
}

public static boolean solve(int[][] sud) {
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            if (sud[i][j] != 0) {
                continue;
            }
            for (int x = 1; x < 10; x++) {
                if (!used(i, j, x)) {
                    sud[i][j] = x;
                    if (solve(sud)) {
                        return true;
                    }
                }
            }
            return false;
        }
    }
    return true;

}

public static boolean isinrow(int i, int j, int x) {
    for (int t = 0; t < 9; t++) {
        if (sud[i][t] == x) {
            return true;
        }
    }
    return false;
}

public static boolean isincol(int i, int j, int x) {
    for (int t = 0; t < 9; t++) {
        if (sud[t][j] == x) {
            return true;
        }
    }
    return false;

}

public static boolean isinsq(int sr, int sc, int x) {
    for (sr = 0; sr < 3; sr++) {
        for (sc = 0; sc < 3; sc++) {
            if (sud[sr][sc] == x) {
                return true;
            }
        }
    }
    return false;
}

static boolean used(int i, int j, int x) {
    if (!isinrow(i, j, x)) {
        if (!isincol(i, j, x)) {
            if (!isinsq(i - (i % 3), j - (j % 3), x)) {
                return false;
            }
        }
    }
    return true;
}

}

Answer 1

你的问题出在你正在执行的这个函数中

 public static boolean isinsq(int sr, int sc, int x) {
    for ( sr = 0; sr < 3; sr++) {
         //  ^ you are reseting the value of sr that you pass in
          //   effectivel making it so that you always check the first square
        for (  sc = 0; sc < 3; sc++) {
               // ^ same here    
            if (sud[sr][sc] == x) {
                return true;
            }
        }
    }
    return false;
}

这会将 sr 重置回 0，因此您只检查了左上角而不是您传入的坐标。您应该这样做:

public static boolean isinsq(int xcorner, int ycorner, int x) {
    for (int sr = xcorner; sr < 3; sr++) {
           //^ here w create a new variable with the starting value that you passed in
        for ( int sc = ycorner; sc < 3; sc++) {
                   //^ here w create a new variable with the starting value that you passed in
            if (sud[sr][sc] == x) {
                return true;
            }
        }
    }
    return false;
}

顺便说一句，解决方案中的双嵌套 for 循环是不必要的，并且会增加大量开销。与其寻找下一个开放点，为什么不假设它们全部开放并检查它们是否不是这样，您的递归将为您处理该迭代。考虑这样的事情(它也更简单......至少对我来说)

bool solve(int[][] sud, int x, int y)
{
//here need to make sure that when x>9 we set x to 0 and increment y
//also if x == 9 and y == 9 need to take care of end condition

if(sud[x][y]==0) //not a pre given value so we are ok to change it
{
    for(int i =1; i<10; ++i)
    {
        sud[x][y] = i;
        if(validBoard(sud)) //no point in recursing further if the current board isnt valid
        {
            if(solve(sud, x+1,y))
            {
                return true;
            }
        }
    }
}
else
{
    return solve(x+1,y);
}
return false;
}

我留下了一些需要填写的地方(让我为你做这件事有什么乐趣:)。正如其他人建议的那样，您应该使用更有意义的变量名称。