最长的夜晚的最佳搜索算法-Javascript
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给定一系列酒店客房及其有效期(1月1日至1月6日):
[
{
roomId: 101,
availability: [
{ roomId: 101, date: '2018-01-01' },
{ roomId: 101, date: '2018-01-02' },
{ roomId: 101, date: '2018-01-03' },
{ roomId: 101, date: '2018-01-05' },
{ roomId: 101, date: '2018-01-06' }
]
},
{
roomId: 102,
availability: [
{ roomId: 102, date: '2018-01-01' },
{ roomId: 102, date: '2018-01-03' },
{ roomId: 102, date: '2018-01-04' },
{ roomId: 102, date: '2018-01-05' }
]
},
{
roomId: 103,
availability: [
{ roomId: 103, date: '2018-01-02' },
{ roomId: 103, date: '2018-01-03' },
{ roomId: 103, date: '2018-01-06' }
]
},
{
roomId: 104,
availability: [
{ roomId: 104, date: '2018-01-04' },
{ roomId: 104, date: '2018-01-05' },
{ roomId: 104, date: '2018-01-06' }
]
},
{
roomId: 105,
availability: [
{ roomId: 105, date: '2018-01-01' },
{ roomId: 105, date: '2018-01-02' },
{ roomId: 105, date: '2018-01-04' },
{ roomId: 105, date: '2018-01-06' }
]
}
]
上面的可用性的表格说明:
| | 1 Jan | 2 Jan | 3 Jan | 4 Jan | 5 Jan | 6 Jan |
| 101 | O | O | O | | O | O |
| 102 | O | | O | O | O | |
| 103 | | O | O | | | O |
| 104 | | | | O | O | O |
| 105 | O | O | | O | | O |
根据上面的输入,预期结果是具有分组可用性的最终会议室:
{
roomId: 101, // determined by the first object in the array
availability: [
{ roomId: 101, date: '2018-01-01' },
{ roomId: 101, date: '2018-01-02' },
{ roomId: 101, date: '2018-01-03' },
{ roomId: 104, date: '2018-01-04' },
{ roomId: 104, date: '2018-01-05' },
{ roomId: 104, date: '2018-01-06' }
]
}
Final grouping selection to be: 101 & 104
| | 1 Jan | 2 Jan | 3 Jan | 4 Jan | 5 Jan | 6 Jan |
| 101 | ?? | ?? | ?? | | O | O |
| 102 | O | | O | O | O | |
| 103 | | O | O | | | O |
| 104 | | | | ?? | ?? | ?? |
| 105 | O | O | | O | | O |
因此,如何根据整个住宿期间最少的房间移动来确定最终选择.
在javascript中执行此操作最有效的搜索算法是什么(就性能而言)? (即使在长期要求可用性或将更多房间分组的情况下,也需要有效地保持真正快速的处理)
I’ll put my algorithm in the answer section, but I don’t think it is
the most efficient way to do it. Please suggest if there is a better way!
解决方法:
我建议在几天中使用循环,并在每次迭代中,确定从当天开始连续营业时间最长的房间;然后,将天数增加该天数.
为了简化解析,您可以对数据进行预处理,以使可用性信息在以数字日期索引索引的对象中可用-例如,将
availability: [
{ roomId: 105, date: '2018-01-01' },
{ roomId: 105, date: '2018-01-02' },
{ roomId: 105, date: '2018-01-04' },
{ roomId: 105, date: '2018-01-06' }
]
进入
'105': {
1: true,
2: true,
4: true,
6: true
}
这样,要弄清楚从第N天起房间X的可用时间,只需重复对room [x] [n] === true的测试,并递增n直到测试失败.
如果这是多次执行(从相同的房间数据集开始),则可以提前一次进行所有实际计算,并构造一个包含每天最佳房间的对象,例如:
{ // keys represent day index
1: { roomId: 101, availableUntil: 3 },
2: { roomId: 101, availableUntil: 3 }, // just as good as room 103
3: { roomId: 101, availableUntil: 3 }, // just as good as 103 and 102
// room 101 not available on Jan 4, room 104 becomes the best room to choose:
4: { roomId: 104, availableUntil: 6 },
5: { roomId: 104, availableUntil: 6 },
6: { roomId: 104, availableUntil: 6 } // just as good as 105
}
然后,在输入了想要停留的天数之后,计算最不具破坏性的房间变化是财产查询的简单问题,直到结束时为止.
为了简化以下代码的可读性,我将使用一个辅助函数
const dateStrToDayIndex = dateStr => Number(dateStr.match(/\d\d$/)[0]);
以便使索引从1开始到6,以测试您的输入,但是在您的真实代码中,您当然应该使用可靠的方法来计算dateStr和某个日期之间的天数,例如1970年1月1日,或者这样的事情. (或者,如果适合的话,可以随意使用您当前正在使用的moment(toDate).diff(moment(fromDate),’days’)
现在,执行代码:首先将数据集转换为如下所示的对象:
/*
{
101: {
1: true,
2: true,
3: true,
5: true,
6: true,
},
102:
// ...
}
*/
const dateStrToDayIndex = dateStr => Number(dateStr.match(/\d\d$/)[0]);
const datasetByRoom = dataset.reduce((datasetA, { roomId, availability }) => {
datasetA[roomId] = availability.reduce((a, { date }) => {
a[dateStrToDayIndex(date)] = true;
return a;
}, {});
return datasetA;
}, {});
然后,主要的getBestRoomFromDay函数获取日期索引,并搜索datasetByRoom中的每个房间对象,以查找从当前日期开始连续营业的对象:
function getBestRoomFromDay(dayIndex) {
let bestRoomSoFar;
let bestCumulativeDaysSoFar = 0;
Object.entries(datasetByRoom).forEach(([room, availObj]) => {
let thisRoomDays = 0;
let dayIndexCheck = dayIndex;
while (availObj[dayIndexCheck]) {
dayIndexCheck++;
thisRoomDays++;
}
if (thisRoomDays > bestCumulativeDaysSoFar) {
bestRoomSoFar = room;
bestCumulativeDaysSoFar = thisRoomDays - 1;
}
});
return {
room: bestRoomSoFar,
until: dayIndex + bestCumulativeDaysSoFar
};
}
在操作中,为输入中的每个dayIndex调用getBestRoomFromDay的示例(1-6):
const dataset=[{roomId:101,availability:[{roomId:101,date:'2018-01-01'},{roomId:101,date:'2018-01-02'},{roomId:101,date:'2018-01-03'},{roomId:101,date:'2018-01-05'},{roomId:101,date:'2018-01-06'}]},{roomId:102,availability:[{roomId:102,date:'2018-01-01'},{roomId:102,date:'2018-01-03'},{roomId:102,date:'2018-01-04'},{roomId:102,date:'2018-01-05'}]},{roomId:103,availability:[{roomId:103,date:'2018-01-02'},{roomId:103,date:'2018-01-03'},{roomId:103,date:'2018-01-06'}]},{roomId:104,availability:[{roomId:104,date:'2018-01-04'},{roomId:104,date:'2018-01-05'},{roomId:104,date:'2018-01-06'}]},{roomId:105,availability:[{roomId:105,date:'2018-01-01'},{roomId:105,date:'2018-01-02'},{roomId:105,date:'2018-01-04'},{roomId:105,date:'2018-01-06'}]}];const dateStrToDayIndex=dateStr=>Number(dateStr.match(/\d\d$/)[0]);const datasetByRoom=dataset.reduce((datasetA,{roomId,availability})=>{datasetA[roomId]=availability.reduce((a,{date})=>{a[dateStrToDayIndex(date)]=!0;return a},{});return datasetA},{});function getBestRoomFromDay(dayIndex){let bestRoomSoFar;let bestCumulativeDaysSoFar=0;Object.entries(datasetByRoom).forEach(([room,availObj])=>{let thisRoomDays=0;let dayIndexCheck=dayIndex;while(availObj[dayIndexCheck]){dayIndexCheck++;thisRoomDays++}
if(thisRoomDays>bestCumulativeDaysSoFar){bestRoomSoFar=room;bestCumulativeDaysSoFar=thisRoomDays-1}});return{room:bestRoomSoFar,until:dayIndex+bestCumulativeDaysSoFar}}
console.log('Example of testing getBestRoomFromDay function on all days:');
for (let i = 1; i < 7; i++) {
console.log('Day ' + i + ': ' + JSON.stringify(getBestRoomFromDay(i)));
}
然后,要构造一个从日期到日期的字符串(例如“ 2018-01-01”至“ 2018-01-06”)的时间表,只需重复调用具有适当日期的getBestRoomFromDay,就可以将每个日期的日索引增加所需的数量迭代:
const dataset=[{roomId:101,availability:[{roomId:101,date:'2018-01-01'},{roomId:101,date:'2018-01-02'},{roomId:101,date:'2018-01-03'},{roomId:101,date:'2018-01-05'},{roomId:101,date:'2018-01-06'}]},{roomId:102,availability:[{roomId:102,date:'2018-01-01'},{roomId:102,date:'2018-01-03'},{roomId:102,date:'2018-01-04'},{roomId:102,date:'2018-01-05'}]},{roomId:103,availability:[{roomId:103,date:'2018-01-02'},{roomId:103,date:'2018-01-03'},{roomId:103,date:'2018-01-06'}]},{roomId:104,availability:[{roomId:104,date:'2018-01-04'},{roomId:104,date:'2018-01-05'},{roomId:104,date:'2018-01-06'}]},{roomId:105,availability:[{roomId:105,date:'2018-01-01'},{roomId:105,date:'2018-01-02'},{roomId:105,date:'2018-01-04'},{roomId:105,date:'2018-01-06'}]}];const dateStrToDayIndex=dateStr=>Number(dateStr.match(/\d\d$/)[0]);const datasetByRoom=dataset.reduce((datasetA,{roomId,availability})=>{datasetA[roomId]=availability.reduce((a,{date})=>{a[dateStrToDayIndex(date)]=!0;return a},{});return datasetA},{});function getBestRoomFromDay(dayIndex){let bestRoomSoFar;let bestCumulativeDaysSoFar=0;Object.entries(datasetByRoom).forEach(([room,availObj])=>{let thisRoomDays=0;let dayIndexCheck=dayIndex;while(availObj[dayIndexCheck]){dayIndexCheck++;thisRoomDays++}
if(thisRoomDays>bestCumulativeDaysSoFar){bestRoomSoFar=room;bestCumulativeDaysSoFar=thisRoomDays-1}});return{room:bestRoomSoFar,until:dayIndex+bestCumulativeDaysSoFar}};
function getSchedule(dateStrFrom, dateStrTo) {
const [from, to] = [dateStrFrom, dateStrTo].map(dateStrToDayIndex);
let day = from;
const schedule = [];
while (day < to) {
const schedObj = getBestRoomFromDay(day);
schedule.push({ from: day, ...schedObj });
// increment day, so as to find the next longest consecutive room:
day = schedObj.until + 1;
}
schedule[schedule.length - 1].until = to;
return schedule;
}
console.log(getSchedule('2018-01-01', '2018-01-06'));
完整,未缩小:
const dataset = [
{
roomId: 101,
availability: [
{ roomId: 101, date: '2018-01-01' },
{ roomId: 101, date: '2018-01-02' },
{ roomId: 101, date: '2018-01-03' },
{ roomId: 101, date: '2018-01-05' },
{ roomId: 101, date: '2018-01-06' }
]
},
{
roomId: 102,
availability: [
{ roomId: 102, date: '2018-01-01' },
{ roomId: 102, date: '2018-01-03' },
{ roomId: 102, date: '2018-01-04' },
{ roomId: 102, date: '2018-01-05' }
]
},
{
roomId: 103,
availability: [
{ roomId: 103, date: '2018-01-02' },
{ roomId: 103, date: '2018-01-03' },
{ roomId: 103, date: '2018-01-06' }
]
},
{
roomId: 104,
availability: [
{ roomId: 104, date: '2018-01-04' },
{ roomId: 104, date: '2018-01-05' },
{ roomId: 104, date: '2018-01-06' }
]
},
{
roomId: 105,
availability: [
{ roomId: 105, date: '2018-01-01' },
{ roomId: 105, date: '2018-01-02' },
{ roomId: 105, date: '2018-01-04' },
{ roomId: 105, date: '2018-01-06' }
]
}
];
const dateStrToDayIndex = dateStr => Number(dateStr.match(/\d\d$/)[0]);
const datasetByRoom = dataset.reduce((datasetA, { roomId, availability }) => {
datasetA[roomId] = availability.reduce((a, { date }) => {
a[dateStrToDayIndex(date)] = true;
return a;
}, {});
return datasetA;
}, {});
function getBestRoomFromDay(dayIndex) {
let bestRoomSoFar;
let bestCumulativeDaysSoFar = 0;
Object.entries(datasetByRoom).forEach(([room, availObj]) => {
let thisRoomDays = 0;
let dayIndexCheck = dayIndex;
while (availObj[dayIndexCheck]) {
dayIndexCheck++;
thisRoomDays++;
}
if (thisRoomDays > bestCumulativeDaysSoFar) {
bestRoomSoFar = room;
bestCumulativeDaysSoFar = thisRoomDays - 1;
}
});
return {
room: bestRoomSoFar,
until: dayIndex + bestCumulativeDaysSoFar
};
}
function getSchedule(dateStrFrom, dateStrTo) {
const [from, to] = [dateStrFrom, dateStrTo].map(dateStrToDayIndex);
let day = from;
const schedule = [];
while (day < to) {
const schedObj = getBestRoomFromDay(day);
schedule.push({ from: day, ...schedObj });
// increment day, so as to find the next longest consecutive room:
day = schedObj.until + 1;
}
schedule[schedule.length - 1].until = to;
return schedule;
}
console.log(getSchedule('2018-01-01', '2018-01-06'));
如前所述,如果必须从同一个数据集计算多个getBestRoomFromDays(即,不插入新的保留),则可以预先构造一个对象,该对象包含可以调用的每个可能值的getBestRoomFromDay,从而确保每天的计算将只进行一次.
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