PHP中数字范围的计算

首先，感谢您抽出时间阅读我的问题

我试图写一个脚本，我遇到了一个我发现很难解决的问题。我正在处理一对数字，例如1000和2000，我有一个数字对数组：

$pairs = array(
    array(800, 1100),
    array(1500, 1600),
    array(1900, 2100)
)

我试图找到的是，如何得到数字对未覆盖的范围，介于1000和2000之间。在本例中，1000-1100由阵列800覆盖，1100、1500-1600由阵列1500、1600覆盖，1900-2000由阵列1900、2100覆盖，剩下1101-1499和1599-1899要覆盖。我希望我说得够清楚了

我想知道的是如何让PHP返回给我一个$pairs变量未包含的范围数组。在本例中，它将返回：

array(
    array(1101, 1499),
    array(1599, 1899)
)

你知道什么是最好的方法吗

---

提前谢谢你。

我会这样做：

begin = 1000
end   = 2000
uncovered = ()
foreach pairs as pair
  if (pair[0] > begin)
    push (uncovered, begin, pair[0])
    begin = pair[1]
  end if
end foreach

这只是一个想法，但关键是： 假设你有一个从1000到2000的小片段。你想得到大的一个没有被小的一个覆盖的每一个部分。想象你有一支钢笔

从头开始。迭代你拥有的每个小片段。如果你严格地在开始之后，那么就有一个洞，所以你必须记住从开始到当前片段的开始

---

希望这有帮助，这是正确的

首先，您必须定义问题：

这两对是按顺序排列的吗？ 成对重叠吗？ 您想查找特定范围的缺失范围似乎是这样吗？ 如果对未排序，请首先对其进行排序：

usort($pairs, 'cmp_pair');

function cmp_pair($a, $b) {
  if ($a[0] == $b[0]) {
    if ($a[1] == $b[1]) {
      return 0;
    } else {
      return $a[1] < $b[1] ? -1 : 1;
    }
  } else {
    return $a[0] < $b[0] ? -1 : 1;
  }
}

现在不应该有任何重叠对，所以问题变得简单了一点，因为您还得到了一个排序数组

function missing($start, $end, $pairs) {
  $missing = array();
  $prev = false;
  foreach ($pairs as $pair) {
    // if the current pair starts above the end, we're done
    if ($pair[0] > $end) {
      break;
    }

    // we can ignore any pairs that end before the start
    if ($pair[1] < $start) {
      continue;
    }

    // if the pair encompasses the whole range, nothing is missing
    if ($pair[0] <= $start && $pair[1] >= $end) {
      break;
    }

    // if this is our first overlapping pair and it starts above
    // the start we can backfill the missing range
    if ($pair[0] > $start && !$missing) {
      $missing[] = array($start, $pair[0]);
    }

    // compare this pair to the previous one (if there is one) and
    // fill in the missing range
    if ($prev) {
      $missing[] = array($prev[1]+1, $pair[0]-1);
    }

    // set the previous
    $prev = $pair;
  }

  // if this never got set the whole range is missing
  if (!$prev) {
    $missing[] = array($start, $end);

  // if the last overlapping range ended before the end then
  // we are missing a range from the end of it to the end of
  // of the relevant range
  } else if ($prev[1] < $end) {
    $missing[] = array($prev[1]+1, $end);
  }

  // done!
  return $missing;
}

希望有帮助

// your original arrays of integers
$pairs = array(
    array(800, 1100),
    array(1500, 1600),
    array(1900, 2100)
);

// first, normalize the whole thing into a giant list of integers that
// are included in the array pairs, combine and sort numerically
$numbers_in_pairs = array();
foreach($pairs as $set) {
    $numbers_in_pairs = array_merge($numbers_in_pairs, range($set[0], $set[1]));
}
sort($numbers_in_pairs);

// find the min
$min = $numbers_in_pairs[0];

// find the max
$max = $numbers_in_pairs[count($numbers_in_pairs)-1];

找出数组的差异

关于我们稍后将使用的集合的元数据：

结果:

---

这个答案对我来说非常有效，谢谢你：我唯一需要更正的是如果$prev。。。如果$pair[0}>$start&&！$missing，则应该在前面，因为对于800-1100，函数只设置$prev，这意味着当它到达第二对时，它仍然认为它是第一对，因此1000-1500之间的所有内容都被视为丢失。非常感谢您的帮助，cletus：

// your original arrays of integers
$pairs = array(
    array(800, 1100),
    array(1500, 1600),
    array(1900, 2100)
);

// first, normalize the whole thing into a giant list of integers that
// are included in the array pairs, combine and sort numerically
$numbers_in_pairs = array();
foreach($pairs as $set) {
    $numbers_in_pairs = array_merge($numbers_in_pairs, range($set[0], $set[1]));
}
sort($numbers_in_pairs);

// find the min
$min = $numbers_in_pairs[0];

// find the max
$max = $numbers_in_pairs[count($numbers_in_pairs)-1];

// create an array of all numbers inclusive between the min and max
$all_numbers = range($min, $max);

// the numbers NOT included in the set can be found by doing array_diff
// between the two arrays, we need to sort this to assure no errors when
// we iterate over it to get the maxes and mins
$not_in_set = array_diff($all_numbers, $numbers_in_pairs);
sort($not_in_set);

// gather metadata about the numbers that are not inside the set
// $not_in_set_meta['min'] = lowest integer
// $not_in_set_meta['max'] = highest integer
// $not_in_set_meta['mins'] = min boundary integer
// $not_in_set_meta['maxes'] = max boundary integer
$not_in_set_meta = array();
for($i=0;$i<count($not_in_set);$i++) {
    if ($i == 0) {
        $not_in_set_meta['min'] = $not_in_set[$i];
        $not_in_set_meta['mins'][] = $not_in_set[$i];
    } else if ($i == count($not_in_set)-1 ) {
        $not_in_set_meta['max'] = $not_in_set[$i];
        $not_in_set_meta['maxes'][] = $not_in_set[$i];
    } else {
        // in the event that a number stands alone
        // that it can be BOTH the min and the max
        if (($not_in_set[$i+1] - $not_in_set[$i]) > 1) {
            $not_in_set_meta['maxes'][] = $not_in_set[$i];
        }
        if (($not_in_set[$i] - $not_in_set[$i-1]) > 1) {
            $not_in_set_meta['mins'][] = $not_in_set[$i];
        }
    }
}

// The final variable which we'll dump the ranges not covered into:
$non_sets = array();

while(count($not_in_set_meta['mins']) > 0 && count($not_in_set_meta['maxes'])) {
    $non_sets[] = array(array_shift($not_in_set_meta['mins']), 
                        array_shift($not_in_set_meta['maxes']));
}
// print it out:
print var_export($non_sets);

array (
  0 => 
  array (
    0 => 1101,
    1 => 1499,
  ),
  1 => 
  array (
    0 => 1601,
    1 => 1899,
  ),
)

?>