如何通过抽取函数和优化循环条件来简化判断连成三子情况的代码？

/// ... 省略代码

// 用于判断是否有连成三子的情况
function determineequare3(maxtime = 3, userplace = 1, box = previousbox) {
  let times = 0;
  let prep = -1;
  let geyi = false;
  let temparr = [];
  // ... 省略代码
}

/// ... 省略代码

// 用于判断是否有连成三子的情况
function determineEquare3(maxTime = 3, userPlace = 1, box = previousBox) {
  return checkDirections(box, userPlace, maxTime);
}

function checkDirections(box, userPlace, maxTime) {
  const directions = [
    [0, 1], [1, 0], [1, 1], [1, -1] // 四个方向：水平、垂直、两个对角线
  ];

  let result;
  for (const [dx, dy] of directions) {
    result = checkDirection(box, userPlace, dx, dy, maxTime);
    if (result) {
      break;
    }
  }

  return result;
}

function checkDirection(box, userPlace, dx, dy, maxTime) {
  let times = 0;
  let preP = -1;
  let geyi = false;
  let x = box.x;
  let y = box.y;

  while (x >= 0 && x < row.value && y >= 0 && y < col.value) {
    const place = boxs.value[x][y].place;
    if (place === userPlace) {
      if (preP === -1 || preP === y - dy) {
        if (preP === y - dy && geyi) {
          x--;
          continue;
        }
        times++;
      } else {
        times = 1;
        geyi = false;
      }
    } else {
      if (times === maxTime - 1 && geyi) {
        x--;
        continue;
      }
      break;
    }

    preP = y;
    x += dx;
    y += dy;
  }

  if (times >= maxTime) {
    return { x, y, times, geyi };
  }

  return false;
}