您可以對數組應用二進制搜索。前提是您的數組已排序。->O(log(n))
[obj1,obj2,obj3....obj100]
測試中間的對象(obj50)，然后決定是否必須在一半[obj1...obj49]或一半中進行搜索，[obj51...obj100]
否則您可以將對象（事件）傳遞到其他數據結構中，例如樹。->O(log(n)) 只是循環遍歷整個數組不會有效率，但如果你不重復它太多也可以。但是從一開始就對數組進行排序將是最好的解決方案。
編輯：以下代碼顯示了二進制搜索實現的基本示例。
const events = [{
eventTitle: "Event title 1",
eventId: "[email protected]",
startDate: "Sun Mar 18 00:00:00 GMT+01:00 2018",
endDate: "Mon Mar 19 00:00:00 GMT+01:00 2018"
},
{
eventTitle: "Event title 2",
eventId: "[email protected]",
startDate: "Tue Mar 19 00:00:00 GMT+01:00 2019",
endDate: "Wed Mar 20 00:00:00 GMT+01:00 2019"
},
{
eventTitle: "Event title 3",
eventId: "[email protected]",
startDate: "Fri Mar 20 00:00:00 GMT+01:00 2020",
endDate: "Sat Mar 21 00:00:00 GMT+01:00 2020"
},
{
eventTitle: "Event title 4",
eventId: "[email protected]",
startDate: "Fri Mar 21 00:00:00 GMT+01:00 2021",
endDate: "Sat Mar 22 00:00:00 GMT+01:00 2021"
},
{
eventTitle: "Event title 5",
eventId: "[email protected]",
startDate: "Fri Mar 22 00:00:00 GMT+01:00 2022",
endDate: "Sat Mar 23 00:00:00 GMT+01:00 2022"
}
];
function binarySearch(array, value, borderLeft, borderRight) {
if (borderLeft <= borderRight) {
var index = Math.floor((borderLeft + borderRight) / 2);
var number = getNumberFromTitle(array[index].eventTitle);
if (number == value) {
return array[index].startDate;
} else if (number > value) {
return binarySearch(array, value, borderLeft, index - 1);
} else {
return binarySearch(array, value, index + 1, borderRight);
}
} else {
return null;
}
}
function getNumberFromTitle(title) {
var tmp = title.split(" ");
return tmp[tmp.length - 1];
}
console.log(binarySearch(events, 4, 0, events.length - 1));
