分割重疊範圍

Question

我已經實現了將重疊現有範圍劃分爲日期/數字範圍列表的算法。

該算法正在工作，但我想知道如果你可以避免算法的最後一個循環。

有可能嗎？

Input Data

0-100（紅） 90-150（綠色） 90-150（藍色） 140-300（黃色） 170-240（黑色） 350-530（橙色） 50-500（銀） 50-60（粉紅色）

Output Data

0-49（紅色） 50-60（紅色，銀色，粉色） 61-89（紅色，銀色）01 （綠色，藍色，銀色） 140-150（綠色，藍色，黃色，銀色） （黃，黑，銀） 241-300（黃色，銀） 301-349（銀） 350-500（橙色，銀色） 501-530（橙色）

JavaScript代碼：

function splitRanges(original_intervals) { 
 
    
 
     for (var to = [], from = [], n, i = original_intervals.length; i--;) { 
 
      if (to.indexOf(n = original_intervals[i].to) < 0) 
 
       to.push(n); 
 
      if (from.indexOf(n = original_intervals[i].from) < 0) 
 
       from.push(n); 
 
     } 
 
    
 
    
 
     to.sort(function(a, b) { 
 
      return a - b; 
 
     }); 
 
     from.sort(function(a, b) { 
 
      return a - b; 
 
     }); 
 
    
 
    
 
    
 
    
 
     var intervals = []; 
 
     while (to.length) { 
 
      var sFrom = from.shift(); 
 
      var sTo = 0; 
 
      if (from.length == 0) { 
 
       sTo = (from.push((n = to.shift()) + 1), n); 
 
    
 
      } else { 
 
       if (from[0] > to[0]) { 
 
        while (to[0] < from[0]) { 
 
    
 
         from.unshift(to[0] + 1); 
 
         to.shift(); 
 
    
 
        } 
 
    
 
        sTo = from[0] - 1; 
 
       } else { 
 
        sTo = from[0] - 1; 
 
    
 
       } 
 
      } 
 
      intervals.push({ 
 
       from: sFrom, 
 
       to: sTo, 
 
       colors: [] 
 
      }); 
 
     } 
 
    
 
     /***********************Loop that i want remove*/ 
 
    
 
     intervals.forEach(function(item, index) { 
 
      original_intervals.forEach(function(item1, index1) { 
 
       if ((item.from >= item1.from && item.from <= item1.to) || (item.to >= item1.from && item.to <= item1.to)) 
 
        item.colors.push(item1.color); 
 
      }); 
 
    
 
     }); 
 
    
 
     return intervals; 
 
    } 
 
    
 
    
 
    
 
    
 
    var r1 = [{ 
 
    id: 1, 
 
    from: 0, 
 
    to: 100, 
 
\t color:'red' 
 
}, { 
 
    id: 2, 
 
    from: 90, 
 
    to: 150, 
 
\t color:'green' 
 
}, { 
 
    id: 3, 
 
    from: 90, 
 
    to: 150, 
 
\t color:'blue' 
 
}, { 
 
    id: 4, 
 
    from: 140, 
 
    to: 300, 
 
\t color:'yellow' 
 
}, { 
 
    id: 5, 
 
    from: 170, 
 
    to: 240, 
 
\t color:'black' 
 
}, { 
 
    id: 6, 
 
    from: 350, 
 
    to: 530, 
 
\t color:'orange' 
 
}, { 
 
    id: 7, 
 
    from: 50, 
 
    to: 500, 
 
\t color:'silver' 
 
} 
 
, { 
 
    id: 8, 
 
    from: 50, 
 
    to: 60, 
 
\t color:'pink' 
 
} 
 

 
]; 
 

 

 
console.log(splitRanges(r1)); 
 
    
 
    
 

Answer 1

您需要進行一些迭代，至少是獲取所有範圍點的迭代，然後從中生成一個數組，併爲每個小間隔獲取顏色的子集。

var data = [{ from: 0, to: 100, color: 'red' }, { from: 90, to: 150, color: 'green' }, { from: 90, to: 150, color: 'blue' }, { from: 140, to: 300, color: 'yellow' }, { from: 170, to: 240, color: 'black' }, { from: 350, to: 530, color: 'orange' }, { from: 50, to: 500, color: 'silver' }, { from: 50, to: 60, color: 'pink' }], 
 
    ranges = new Set, 
 
    parts, 
 
    result; 
 

 
data.forEach(({ from, to }) => (ranges.add(from), ranges.add(to))); 
 
parts = [...ranges].sort((a, b) => a - b); 
 
result = parts.slice(1).map(function (a, i, aa) { 
 
    var from = i ? aa[i - 1] : parts[0], 
 
     to = a, 
 
     colors = data.filter(d => d.from <= from && to <= d.to).map(({ color }) => color); 
 
    return { from, to, colors }; 
 
}); 
 
    
 
console.log(result);

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