找到一个数字范围内的差距

Question

I am looking for a solution (preferably JavaScript) that would find gaps in a select set of child ranges compared to their parent range.

Example 1: If I select a parent range of {1,10} and child ranges of {1,2} and {2,10} I would have a gap of 0, meaning the range in the parent range has been completely filled by it's children.

Example 2: If I select a parent range of {1,10} and child ranges of {1-3} and {6,8} I would have a gap of 4.

Answer 1

1. Sort the child ranges by their initial number
2. Subtract the initial number of the parent range from the initial number of the first child range. This is the start of your running total.
3. Subtract the second number of a range from the initial number of the following range
4. If the result of 3. is positive, add to the running total
5. Finally subtract the second number of the last range from the second number of the parent range, add to running total

The resulting total is the number of missing numbers, assuming you're only using integers.

Answer 2

In general the gap will also be a range or set of ranges. Example parent: {1-10}, children {1,2},{5,8}, gap {3,4},{9,10} because of this I would just suggest you write a thing which can subtract one range from another then apply it to the parent for each child. There are 3 cases to consider: is it at the start, in the middle (generating two ranges) or at the end. Then when you are subtracting from a set of ranges you have to consider all cases where there could be overlap.

so in javascript make a range object with a start and end property then carry around arrays of these. The make a function rangeSub(parent, child) to do the subtraction where parent can be an array of ranges and child is a single range.

rangeSub(parent, child) {
   var result;
   //code for set of range subtraction
   if(parent.length)
      for(range in parent) {
          temp = rangeSub(range,child);
          if(temp.length) result.concat(temp);
          else result.push(temp); 
      }
      return result;
   }
   //code for single range subtraction
   if(parent.start < child.start) {
      ...
   }
   if(parent.end > child.end) {
      ...
   }
   etc.
}

There are still several edge cases to work out but this is the general form I would follow.