javascript - 如何合并间距数组

Question

有一个区间数组，需要组合重叠，即:

[[0, 33], [66, 80]] => [[0, 33], [66, 80]]

[[0, 33], [66, 80], [0, 66], [33, 100]] => [0,100]

写代码。我得到结果 [[0, 100], [66, 80]]。这是因为首先范围 [[0, 33]], 然后 [[0, 33], [66, 80]]， 然后 [[0, 66], [66, 80]]， 然后 [[0, 100] , [66, 80]] 好像在不循环的情况下检查其他范围

const data = [
    [0, 33],
    [66, 80],
    [0, 66],
    [33, 100]
  ];

  createDataForSlider = data =>
    data.reduce((prevVal, time) => {
      let isPrevValUpdated = false;

      const timeStart = time[0];
      const timeEnd = time[1];

      /* eslint no-param-reassign: ["error", { "ignorePropertyModificationsFor": ["prevVal"] }] */
      if (prevVal.length) {
        for (let i = 0, ii = prevVal.length; i < ii; i += 1) {
          let prevValCurrent = prevVal[i];

          if (timeStart >= prevValCurrent[0] && timeEnd <= prevValCurrent[1]) {
            isPrevValUpdated = true;
            break;
          }
          if (
            !isPrevValUpdated &&
            timeStart >= prevValCurrent[0] &&
            timeStart <= prevValCurrent[1]
          ) {
            prevValCurrent[1] = timeEnd;
            isPrevValUpdated = true;
            break;
          }
          if (
            !isPrevValUpdated &&
            timeEnd >= prevValCurrent[0] &&
            timeEnd <= prevValCurrent[1]
          ) {
            prevValCurrent[0] = timeStart;
            isPrevValUpdated = true;
            break;
          }

          //console.log("prevVal-", prevVal);
        }
      }
      if (!isPrevValUpdated) {
        prevVal.push([timeStart, timeEnd]);
      }
      return prevVal;
    }, []);

  createDataForSlider(data);

Answer 1

尝试跟随

1. 根据所有开始时间对数组进行排序

2. 现在，通过将先前插入的频率与当前频率进行比较，使用空数组累加器对其进行归约。

   const frequencies = [[0, 33], [66, 80], [0, 66], [33, 100]];
   
   const sortedFrequencies = frequencies.sort((f1, f2) =>{ return f1[0] - f2[0]});
   
   sortedFrequencies.reduce((acc, current) => {
      if(acc.length > 0 && current[0] <= acc[acc.length - 1][1]) {
   
           acc[acc.length - 1][1] = current[1] > acc[acc.length - 1][1] ? current[1] : acc[acc.length - 1][1];
      } else {
   
           acc.push(current)
      }
      return acc;
   }, []);