关于数组拆分的问题？

Question

像这样的数组，如何才能拆分为结果图所示的那样，将时间不连贯的拆开为单独的一项

结果图

Answer 1

const result = {};
const arr = [
    { time: '10:00' },
    { time: '11:00' },
    { time: '12:00' },
    { time: '13:00' },
    { time: '17:00' },
    { time: '18:00' },
    { time: '19:00' },
    { time: '20:00' },
    { time: '21:00' },
    { time: '23:00' },
    { time: '22:00' },
];
const splitNum = 1; //间隔超过1个小时就认为不连续
let partNum = 1;

arr.sort((a, b) => parseInt(a.time) - parseInt(b.time)).map((value, index) => {
    if (index === 0) {
        result[`part_${partNum}`] = [value];
        return
    }
    if (parseInt(value.time) - parseInt(arr[index - 1].time) > splitNum) {//不连续
        partNum += 1;
        result[`part_${partNum}`] = [];
    }

    result[`part_${partNum}`].push(value);
})

console.log(result);

// {
//     part_1: [
//       { time: '10:00' },
//       { time: '11:00' },
//       { time: '12:00' },
//       { time: '13:00' }
//     ],
//     part_2: [
//       { time: '17:00' },
//       { time: '18:00' },
//       { time: '19:00' },
//       { time: '20:00' },
//       { time: '21:00' },
//       { time: '22:00' },
//       { time: '23:00' }
//     ]
//   }

Answer 2

const foo = (res) => res
.map((cur) => Number(cur.time.split(':')[0]))
.reduce((acc, curTime, i, src) => {
  const cur = res[i];

  // 与上一个元素不衔接
  if (!src[i - 1] || curTime - 1 > src[i - 1]) {
    acc.push([cur]);
  } else {
    acc[acc.length - 1].push(cur);
  }

  return acc;
}, []);

// 测试
> foo([{ time: '10:00' }, { time: '11:00' }, { time: '12:00' }, { time: '13:00' }, { time: '17:00' }, { time: '18:00' }, { time: '20:00' }, { time: '22:00' }, { time: '23:00' }]);
< [
    [{ time: "10:00" }, { time: "11:00" }, { time: "12:00" }, { time: "13:00" }],
    [{ time: "17:00" }, { time: "18:00" }],
    [{ time: "20:00" }],
    [{ time: "22:00" }, { time: "23:00" }]
  ];

和上一次答的有那么一些神似。