LeetCode0037解數獨

上篇博文我們認識了什麼是有效的數獨
LeetCode0036有效的數獨

今天我們來解數獨

題目描述

回溯法

使用的概念

瞭解兩個編程概念會對接下來的分析有幫助。

• 第一個叫做 約束編程。

基本的意思是在放置每個數字時都設置約束。在數獨上放置一個數字後立即 排除當前 行， 列和子方塊 對該數字的使用。這會傳播約束條件並有利於減少需要考慮組合的個數。

• 第二個叫做 回溯。

讓我們想象一下已經成功放置了幾個數字在數獨上。
但是該組合不是最優的並且不能繼續放置數字了。該怎麼辦？ 回溯。
意思是回退，來改變之前放置的數字並且繼續嘗試。如果還是不行，再次回溯。

class Solution {
  // box size
  int n = 3;
  // row size
  int N = n * n;

  int [][] rows = new int[N][N + 1];
  int [][] columns = new int[N][N + 1];
  int [][] boxes = new int[N][N + 1];

  char[][] board;

  boolean sudokuSolved = false;

  public boolean couldPlace(int d, int row, int col) {
    /*
    Check if one could place a number d in (row, col) cell
    */
    int idx = (row / n ) * n + col / n;
    return rows[row][d] + columns[col][d] + boxes[idx][d] == 0;
  }

  public void placeNumber(int d, int row, int col) {
    /*
    Place a number d in (row, col) cell
    */
    int idx = (row / n ) * n + col / n;

    rows[row][d]++;
    columns[col][d]++;
    boxes[idx][d]++;
    board[row][col] = (char)(d + '0');
  }

  public void removeNumber(int d, int row, int col) {
    /*
    Remove a number which didn't lead to a solution
    */
    int idx = (row / n ) * n + col / n;
    rows[row][d]--;
    columns[col][d]--;
    boxes[idx][d]--;
    board[row][col] = '.';
  }

  public void placeNextNumbers(int row, int col) {
    /*
    Call backtrack function in recursion
    to continue to place numbers
    till the moment we have a solution
    */
    // if we're in the last cell
    // that means we have the solution
    if ((col == N - 1) && (row == N - 1)) {
      sudokuSolved = true;
    }
    // if not yet
    else {
      // if we're in the end of the row
      // go to the next row
      if (col == N - 1) backtrack(row + 1, 0);
        // go to the next column
      else backtrack(row, col + 1);
    }
  }

  public void backtrack(int row, int col) {
    /*
    Backtracking
    */
    // if the cell is empty
    if (board[row][col] == '.') {
      // iterate over all numbers from 1 to 9
      for (int d = 1; d < 10; d++) {
        if (couldPlace(d, row, col)) {
          placeNumber(d, row, col);
          placeNextNumbers(row, col);
          // if sudoku is solved, there is no need to backtrack
          // since the single unique solution is promised
          if (!sudokuSolved) removeNumber(d, row, col);
        }
      }
    }
    else placeNextNumbers(row, col);
  }

  public void solveSudoku(char[][] board) {
    this.board = board;

    // init rows, columns and boxes
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        char num = board[i][j];
        if (num != '.') {
          int d = Character.getNumericValue(num);
          placeNumber(d, i, j);
        }
      }
    }
    backtrack(0, 0);
  }
}