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PHP中数字范围的计算

  •  3
  • Bruno De Barros  · 技术社区  · 16 年前

    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获得数字对未覆盖的范围,介于1000和2000之间。在本例中,1000-1100由数组(8001100)覆盖,1500-1600由数组(15001600)覆盖,1900-2000由数组(1900,2100)覆盖,剩下1101-1499和1599-1899。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?

    先谢谢你。

    3 回复  |  直到 16 年前
        1
  •  3
  •   cletus    16 年前

    1. 这两对是按顺序排列的吗?
    2. 成对重叠吗?

    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;
      }
    }
    

    如果允许重叠范围,请将对列表转换为非重叠集。关于如何做到这一点,这里有一个建议:

    $prev = false;
    $newpairs = array();
    foreach ($pairs as $pair) {
      if ($prev) {
        // this also handles the case of merging two ranges
        // eg 100-199 with 200 to 250 to 100-250
        if ($prev[1] >= $pair[0]-1) {
          $prev = array($prev[0], max($prev[1], $pair[1]));
        } else {
          $newpairs[] = $prev;
        }
      }
      $prev = $pair;
    }
    $pairs = $newpairs;
    

    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。

        2
  •  0
  •   Aif    16 年前

    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: 假设你有一个大片段(从1000到2000)和一个小片段。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”。如果你在(严格地)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!

        3
  •  0
  •   artlung    16 年前
    // 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);
    

    关于我们稍后将使用的集合的元数据:

    // 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];
            }
        }
    }
    

    最终输出:

    // 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);
    

    结果:

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