在 Javascript 中点击某个键之前计算对象在一行中的出现次数

Question

我有一个数组，看起来像这样：

    var grid = [
        [0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0],
        [0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0],
        [0,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,1,1,0,0,0,0,0]
    ];

我想计算连续 1 的数量，直到我们达到零为止。然后，如果我们再次命中 1，则再次计算连续出现的次数。

So for grid[0], it should return 9.
For grid[1], it should return [3,3].
grid[2], [2,1,1,2].

作为一个额外的乐趣，我也在尝试为“列”做这件事。 IE。 “1”出现在第 1 列的次数，包括 grid[0][1]、grid[1][1]、grid[2][1] 等。如果有更好的方法来组织数据来实现这一点，我愿意接受建议，因为我总共有 25 行/数组在数组中。

我不确定是否有不涉及反复循环数据的方法。目前我正在这样做：

    var guides = [];

    for ( var i = 0; i < grid.length; i++ ) {
      var row = grid[i];
      var chunks = [];
      var count = 0;
    
      for (var j = 0; j < row.length; j++ ) {
         if ( j === 1 ) {
             count++;
         } else {
           chunks.push(count);
           count = 0;
         }
      }

    guides.push(chunks);

    }
    

Answer 1

这是一个示例，您可以如何使用递归来检查那些 1 号罢工。对于列，您可以转置原始 Grid 并重新使用相同的功能。

var grid = [
  [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
]

// Returns the array of 1 strikes in the given arr
const occurrenceOfOne = (arr, strike = 0, final = []) => {
  // Recursive calls until the array is empty
  if (arr.length === 0) {
    return strike === 0 ? final : final.concat([strike])
  }

  // If the item is 0, the strike ends.
  if (arr[0] === 0) {
    if (strike !== 0) {
      return occurrenceOfOne(arr.slice(1), 0, final.concat([strike]))
    }
  }

  // If the item is 1, the strike continues
  if (arr[0] === 1) {
    return occurrenceOfOne(arr.slice(1), strike + 1, final)
  }

  // Default value 0 found and strike is 0 as well.
  return occurrenceOfOne(arr.slice(1), 0, final)
}

// Copied transpose function from this gist:
// https://gist.github.com/femto113/1784503
function transpose(a) {
  return a[0].map((_, c) => a.map((r) => r[c]))
}

console.log(occurrenceOfOne(grid[0]))
console.log(occurrenceOfOne(grid[1]))
console.log(occurrenceOfOne(grid[2]))

const transposedGrid = transpose(grid)
console.log(occurrenceOfOne(transposedGrid[6]))
console.log(occurrenceOfOne(transposedGrid[7]))