java - Android how to stop recursive function when I got one result -

i new android. writing simple app user solve sudoku. user asked input numbers program provide 1 solution puzzle.

my recursive function find results works dont know how stop function when got first solution. therefore, app keep running , crash.

here code

public class solver extends actionbaractivity implements onclicklistener{  private final string tag = "solver"; public final int[][] = new int[9][9]; private solverview solverview; public static final int[][] temp = {     {1,1,1,2,2,2,3,3,3},     {1,1,1,2,2,2,3,3,3},     {1,1,1,2,2,2,3,3,3},     {4,4,4,5,5,5,6,6,6},     {4,4,4,5,5,5,6,6,6},     {4,4,4,5,5,5,6,6,6},     {7,7,7,8,8,8,9,9,9},     {7,7,7,8,8,8,9,9,9},     {7,7,7,8,8,8,9,9,9} };  @override protected void oncreate(bundle savedinstancestate){     super.oncreate(savedinstancestate);     solverview = new solverview(this);     setcontentview(r.layout.activity_solver);     solverview.requestfocus();     solverview.bringtofront();     setcontentview(solverview);  }  @override public boolean oncreateoptionsmenu(menu menu) {     getmenuinflater().inflate(r.menu.solver, menu);     return true; }  @override public boolean onoptionsitemselected(menuitem item) {     int id = item.getitemid();     if (id == r.id.action_settings) {         log.d(tag, "menuselect: something");         run(a, 0);         return true;     }     return super.onoptionsitemselected(item); }  @override public void onclick(view v) {     // todo auto-generated method stub     if (v.getid() == r.id.solve) {         log.d(tag, "solve button pressed");     } }  protected void showkeypad(int x, int y) {     dialog dialog = new keyboard(this, this.solverview);     dialog.show(); }  public boolean setcellifvalid(int selu, int selv, int t) {     log.d(tag, " in setcellifvalid " + selu + " " + selv + " " + t);     this.a[selu][selv] = t;     solverview.invalidate();     return true; }  public string getcellstring(int i, int j) {     int v = this.a[i][j];      if (v == 0)         return "";     else         return string.valueof(v);  }  public static void run(int[][] a, int n) {     if (n >= 81) {         printarray(a);     // want program stop here         return;     }     int = n / 9;     int j = n % 9;     if (a[i][j] != 0) run(a, n + 1);     (int k = 1; k <= 9; k++)          if (valid(a, i, j, k)) {             a[i][j] = k;             run(a, n + 1);             a[i][j] = 0;         } }  public static boolean valid(int[][] a, int u, int v, int k) {     //check column     (int = 0; < 9; i++)         if (a[i][v] == k) return false;     //check row     (int j = 0; j < 9; j++)         if (a[u][j] == k) return false;     //check small square     (int = 0; < 9; i++)         (int j = 0; j < 9; j++)             if (temp[i][j] == temp[u][v] && a[i][j] == k)                 return false;     return true;  }  public static void printarray(int[][] a) {     (int = 0; < a.length; i++) {         (int j = 0; j < a[0].length; j++)             system.out.print(a[i][j] + " ");         system.out.println();     }  } 

}

anyone me? thank reading. sincerely

you have 2 possible approaches choose from. best discard recursive approach , try iterative one. way can stop searching answers when find answer. doesn't need iterative function per se: tail-recursive well.

the easier solution store global variable, boolean solutionfound, , initialize false. then, when 1 of recursions find solution, change flag true. recursive function should check flag, , return if it's true.

this pseudocode of sorts find myself using when ever made brute-force sudoku solver:

void solvesquare(currentlayout, row, column) {     if row > maxrows         print(layout);         return;      if column > maxcolumns         solvesquare(currentlayout, row + 1, 0)         return;      foreach digit d in [1, 2, 3, 4, 5, 6, 7, 8, 9] :         if isvaliddigitatsquare(currentlayout, row, column)             temp = currentlayout[row][column];             currentlayout[row][column] = d;             solvesquare(currentlayout, row, column + 1);             currentlayout[row][column] = temp; } 

the above pseudocode assumes there exists 1 solution sudoku, since decent (some prefer valid) sudoku there exists 1 solution. particular pseudocode, have ever return 1 solution add flag follows:

boolean solutionfound = false;  void solvesquare(currentlayout, row, column) {     if solutionfound         return;      if row > maxrows         print(layout);         solutionfound = true;         return;      if column > maxcolumns         solvesquare(currentlayout, row + 1, 0)         return;      foreach digit d in [1, 2, 3, 4, 5, 6, 7, 8, 9] :         if isvaliddigitatsquare(currentlayout, row, column)             temp = currentlayout[row][column];             currentlayout[row][column] = d;             solvesquare(currentlayout, row, column + 1);             currentlayout[row][column] = temp; }