The eight queens problem is solved by javascript recursively

The following is a retrospective solution to the 8 queens, with detailed notes, here is no nonsense.

function NQueens(order) {
  if (order < 4) {
    console.log('N Queens problem apply for order bigger than 3 ! ');
    return;
  }

  var nQueens = [];
  var backTracking = false;
  rowLoop:
    for (var row=0; row<order; row++) {
      // If appear row Less than 0,  There is no solution 
      if(row < 0){
        console.log('This N Queens problem has no solution ! ');
        break;
      }
      // The first 1 A new one was detected this time 1 line 
      if (nQueens[row] === undefined) {
        nQueens[row] = [];
      }
      // A block of programs that run while backtracking 
      for (var col=0; col<order; col++) {
        //0 For the position that has been tested and can place the queen 
        if (nQueens[row][col] === 0) {
          continue;
        }
        // In the process of backtracking, the position where the queen can be put is encountered, which means that the verification of this position in the future has not passed and needs to be processed again 
        else if (backTracking && nQueens[row][col] == 1) {
          // Back in time , on 1 The end of the line , I need to keep going back 
          if (col === order-1) {
            resetRow(nQueens, order, row);
            row = row - 2;
            continue rowLoop;
          }
          // The backtracking line is not at the end of the line ,  mark 0,  Continue to 
          nQueens[row][col] = 0;
          backTracking = false;
          continue;
        }
        // Place the 1 A queen 
        nQueens[row][col] = 1;
        // find 1 The queen can be placed in the position, jump down 1 Line ( 1 Only on the line 1 A queen). 
        if (isQueenValid(nQueens, row, col)) {   
          continue rowLoop;
        }
        // every 1 There should be 1 The queen, to the end of the column has not found a suitable position, indicating that the queen before the placement of a problem, need to backtrack! 
        else if (col == order-1) {       
          backTracking = true;
          //0 with 1 All indicate that this location has been detected, so clear the line as undefined
          resetRow(nQueens, order, row);
          // Reduction of 2 Because the end of the loop has a self addition, which is essentially going back up 1 line 
          row = row - 2;
          // Step back into the outer loop and continue 
          continue rowLoop;           
        } else {
          // If the line is not reached, continue to detect the undetected 
          nQueens[row][col] = 0;        
          continue;
        };
      }
    }
  return nQueens;
}
// Back before ,  Clear line 
function resetRow(nQueens, order, row) {
  for (var col=0; col<order; col++) {
    nQueens[row][col] = undefined;
  }
}
// Check if the position can place the queen 
function isQueenValid(nQueens, row, col) {
  // Line detection 
  for (var i=0; i<col; i++) {
    if (nQueens[row][i] == 1) {
      return false;
    }
  }
  for (var j=1; j<row+1; j++) {
    //    Column tests             Upper left 45 The degree of               The upper right 45 The degree of 
    if (nQueens[row-j][col]==1 || (nQueens[row-j][col-j]==1) || (nQueens[row-j][col+j]==1)) {
      return false;
    }
  }
  return true;
}

function printQ(queens) {
  for (var row=0; row<queens.length; row++) {
    var rowText = '';
    for (var col=0; col<queens.length; col++) {
      if (queens[row][col]===undefined) {
        queens[row][col] = 0;
      }
      rowText = rowText + queens[row][col] + ' ';
    }
    console.log(rowText);
  }
}

var queens = NQueens(8);
printQ(queens);

That's all for this article, I hope you enjoy it.