groovy - 从范围列表中删除重叠范围 Groovy

标签 groovy overlapping

我需要编写一段代码,其中包含 Groovy 中的范围列表。我需要创建一个新列表,其中所有范围都不重叠。

例如,如果输入为:[13..15 , 14..16]

我应该能够创建一个包含 [13..16] 或 [13..14, 14..16] 的列表

我真的很感激任何帮助。我现在已经编写了以下代码,但它一点也不工作:

def removeOverlapInRanges(ranges)
{
    def cleanedRanges = []
    def overLapFound = false
    def rangeIsClean = true
    def test = "ranges"
    ranges.each 
    {
        range->

        def index = ranges.indexOf(range)
        while (index < ranges.size() -1)
        {
            if (ranges.get(index + 1).disjoint(range) == false)
            {
                overLapFound = true
                rangeIsClean = false
                def nextRange = ranges.get(index + 1)
                if (range.from > nextRange.from && range.to < nextRange.to)
                    cleanedRanges.add(range.from..range.to)
                else if (range.from < nextRange.from && range.to < nextRange.to)
                    cleanedRanges.add(range.from..nextRange.to)
                else if (range.from > nextRange.from && range.to > nextRange.to)
                    cleanedRanges.add(nextRange.from..range.to)
            }
            index = index + 1
        }
        if (rangeIsClean)
            cleanedRanges.add(range)

        rangeIsClean = true

        test = test + cleanedRanges
    }
    cleanedRanges.add(0, cleanedRanges.get(cleanedRanges.size()-1))
    cleanedRanges.remove(cleanedRanges.size() - 1)
    if (overLapFound)
        return removeOverlapInRanges(cleanedRanges)
    else
        return cleanedRanges
}

我通过了[12..13, 17..19, 18..22,17..19, 22..23,19..20]

作为返回,我得到了 [12..13]

预先感谢您的任何意见!!

最佳答案

我得到了这个:

List<Range> simplify( List<Range> ranges ) {
  ranges.drop( 1 ).inject( ranges.take( 1 ) ) { r, curr ->
    // Find an overlapping range
    def ov = r.find { curr.from <= it.to && curr.to >= it.from }
    if( ov ) {
      ov.from = [ curr.from, ov.from ].min()
      ov.to   = [ curr.to, ov.to ].max()
      simplify( r )
    }
    else {
      r << curr
    }
  }
}

def ranges = [ 12..13, 17..19, 18..22, 17..19, 22..23, 19..20 ]
assert simplify( ranges ) == [ 12..13, 17..23 ]

ranges = [ -2..3, -5..-2 ]
assert simplify( ranges ) == [ -5..3 ]

ranges = [ 3..1, 1..5 ]
assert simplify( ranges ) == [ 5..1 ] // reversed as first range is reversed

ranges = [ 1..5, 3..1 ]
assert simplify( ranges ) == [ 1..5 ]

ranges = [ 1..5, 3..1, -1..-4 ]
assert simplify( ranges ) == [ 1..5, -1..-4 ]

ranges = [ 1..5, -6..-4, 3..1, -1..-4 ]
assert simplify( ranges ) == [ 1..5, -6..-1 ]

ranges = [ 1..3, 5..6, 3..5 ]
assert simplify( ranges ) == [ 1..6 ]

虽然可能存在边缘情况...所以我会做更多测试...

关于groovy - 从范围列表中删除重叠范围 Groovy,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15135399/

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