拆分日期范围重叠成块（javascript）

Question

我想出了一个概念，但无法将其转换为代码。

这里有一张图片可以更好地帮助解释我的想法：

无法通过图片判断，但日期之间也可能存在间隔。

不要让它成为一个 XY 问题，这就是我需要这样做的原因。我有一系列日期范围和奖金，我想创建一个包含奖金总和的非重叠日期数组。

我正在尝试将重叠部分拆分为不再重叠的卡盘。遇到一个名为moment-range 的包。我想要成功的东西可能有一个名字吗？ “将日期范围投影到线性...某物上”

下面的最终代码

const _ = require('lodash')
const Moment = require('moment')
const MomentRange = require('moment-range')

const moment = MomentRange.extendMoment(Moment);

let bonuses = [
  {start: new Date('Jan 15, 2018 23:00:00 GMT+0200'), end: new Date('Jan 18, 2018 23:59:59 GMT+0200'), bonus: 30, preSale: true},
  {start: new Date('Jan 17, 2018 00:00:00 GMT+0200'), end: new Date('Jan 29, 2018 13:00:00 GMT+0200'), bonus: 25, preSale: true},
  {start: new Date('Feb 12, 2018 00:00:00 GMT+0200'), end: new Date('Feb 18, 2018 23:59:59 GMT+0200'), bonus: 20, preSale: false},
  {start: new Date('Feb 19, 2018 00:00:00 GMT+0200'), end: new Date('Feb 27, 2018 23:59:59 GMT+0200'), bonus: 15, preSale: false},
  {start: new Date('Feb 26, 2018 00:00:00 GMT+0200'), end: new Date('Mar 4, 2018 23:59:59 GMT+0200'), bonus: 10, preSale: false},
  {start: new Date('Mar 5, 2018 00:00:00 GMT+0200'), end: new Date('Mar 11, 2018 23:59:59 GMT+0200'), bonus: 5, preSale: false},
]

_.map(bonuses, o => o.range = moment.range(o.start, o.end))

let dates = []
_.each(bonuses, o => {
  dates.push(o.start)
  dates.push(o.end)
})

dates.sort(function(a, b) {
  return a-b
})

let ranges = []
for(let i=1; i<dates.length; i++) {
  ranges.push({
    start: dates[i-1],
    end: dates[i],
    range: moment.range(dates[i-1], dates[i])
  })
}

for (let range of ranges) {
  range.bonus = 0
  for (let bonus of bonuses) {
    if (range.range.intersect(bonus.range)) {
      range.bonus += bonus.bonus
    }
  }
}

_.each(ranges, r => console.log(r.start, ' - ', r.end, '  => ', r.bonus))

Answer 1

var dates = [];
bonuses.forEach(item => {
    dates.push(item.start);
    dates.push(item.end);
});
dates.sort();
var ranges = [];
for(var i=1;i<dates.length;i++){
    ranges.push({
        start: dates[i-1], 
        end: dates[i]
    });
}
console.log(ranges);

Answer 2

这样的？

var min = Infinity;
var max = 0;
var bonuses = [
    { start: new Date('Jan 15, 2018 23:00:00 GMT+0200'), end: new Date('Jan 18, 2018 23:59:59 GMT+0200'), bonus: 30, intersects: false },
    { start: new Date('Jan 16, 2018 00:00:00 GMT+0200'), end: new Date('Jan 29, 2018 13:00:00 GMT+0200'), bonus: 25, intersects: false },
    { start: new Date('Feb 12, 2018 00:00:00 GMT+0200'), end: new Date('Feb 18, 2018 23:59:59 GMT+0200'), bonus: 20, intersects: false },
    { start: new Date('Feb 19, 2018 00:00:00 GMT+0200'), end: new Date('Feb 29, 2018 23:59:59 GMT+0200'), bonus: 15, intersects: false },
    { start: new Date('Feb 26, 2018 00:00:00 GMT+0200'), end: new Date('Mar 4, 2018 23:59:59 GMT+0200'), bonus: 10, intersects: false },
    { start: new Date('Mar 5, 2018 00:00:00 GMT+0200'), end: new Date('Mar 11, 2018 23:59:59 GMT+0200'), bonus: 5, intersects: false }
];
bonuses
    .map(function (a, index) {
    if (a.start.getTime() < min) {
        min = a.start.getTime();
    }
    if (a.end.getTime() < min) {
        min = a.end.getTime();
    }
    if (a.start.getTime() > max) {
        max = a.start.getTime();
    }
    if (a.end.getTime() > max) {
        max = a.end.getTime();
    }
    for (var i = index + 1; i < bonuses.length; i++) {
        var b = bonuses[i];
        if (b.start.getTime() < a.start.getTime() && b.end.getTime() > a.start.getTime() ||
            a.start.getTime() < b.start.getTime() && a.end.getTime() > b.start.getTime()) {
            bonuses[i].intersects = true;
            a.intersects = true;
        }
    }
    return a;
});
var tileSize = 20;
var canvas = document.body.appendChild(document.createElement("canvas"));
canvas.width = 800;
canvas.height = bonuses.length * tileSize;
var ctx = canvas.getContext("2d");
bonuses.forEach(function (bonus, index) {
    var start = (bonus.start.getTime() - min) / (max - min);
    var end = (bonus.end.getTime() - min) / (max - min);
    ctx.fillStyle = bonus.intersects ? "red" : "green";
    ctx.fillRect(canvas.width * start, tileSize * index, canvas.width * (end - start), tileSize);
    ctx.strokeRect(canvas.width * start, tileSize * index, canvas.width * (end - start), tileSize);
});