转换多个开放时间和减少单个开放时间的问题

Question

我试图解决这个问题。

问题：

给定一个天数组和一个包含开始和结束时间的数据数组 一周内一家商店，查找商店营业的日期/时间。

const daysArr = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday'];
// This is the times for *ONE STORE* in the entire week.
const data = [
{
startTime: '10:00am',
endTime: '5:00pm',
open: [true, true, false, false, false, false, false]
},
{
startTime: '10:00am',
endTime: '5:00pm',
open: [false, false, true, false, false, false, false]
}, 
{
startTime: '11:00am',
endTime: '6:00pm',
open: [false, false, false, true, false, false, true]
} ];

解决方案应该是：

[ "周一至周三上午 10：00 至下午 5：00

"，"周四上午 11：00 至下午 6：00"， "周日上午11：00至下午6：00"]

注意：可能存在差距。例如，商店可能只在星期四和星期日营业(就像在示例中一样(，在这种情况下，只返回它们营业的天数。

我的解决方案：

const daysArr = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday'];
const data = [
{
startTime: '10:00am',
endTime: '5:00pm',
open: [true, true, false, false, false, false, false]
},
{
startTime: '10:00am',
endTime: '5:00pm',
open: [false, false, true, false, false, false, false]
}, 
{
startTime: '11:00am',
endTime: '6:00pm',
open: [false, false, false, true, false, false, true]
}
];
function openStoreTimes(data) {
const store = {};
buildStore(store, data);
return printResult(store, data);
function printResult(store, data) {
const values = Object.keys(store);
const res = [];
values.forEach(key => {
const value = store[key]; // [true, true, true]
res.push(...makeString(value, key));
})
return res;
}
function makeString(arr, key) {
const res = [];
let firstTrueidx=undefined;
for(let i=0; i<arr.length; i++) {
const isTrue = arr[i] === true ? true: false;

if(isTrue) {
if(firstTrueidx === undefined) firstTrueidx = i;
if(arr[i+1] === false || !arr[i+1]) {
let resultKey;
if(firstTrueidx === i) {
resultKey = `${key} ${daysArr[firstTrueidx]}`;
} else {
resultKey = `${key} ${daysArr[firstTrueidx]}-${daysArr[i]}`;
}
res.push(resultKey);
firstTrueidx = undefined;
}
}
}
return res;
}
function buildStore(store, data) {
data.forEach(each => {
const {startTime, endTime, open} = each;
const key = `${startTime}-${endTime}`;
if(!store[key]) {
store[key] = [...open];
} else {
makeTrueArray(store[key], open);
}
});
}
function makeTrueArray(a1, a2) {
a2.forEach((each,i) => {
if(each) a1[i] = true;
});
}
}
console.log(openStoreTimes(data));
// [ '10:00am-5:00pm Monday-Wednesday', '11:00am-6:00pm Thursday', '11:00am-6:00pm Sunday' ]

问题：

我想出了以下解决方案，但是，我觉得这可以简化(我觉得我的解决方案过于复杂(。寻找解决此问题的巧妙方法。

Answer 1

我希望这是实现此问题解决方案的更简洁的方法 -

const daysArr = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday'];
const data = [
{
startTime: '10:00am',
endTime: '5:00pm',
open: [true, true, false, false, false, false, false]
},
{
startTime: '10:00am',
endTime: '5:00pm',
open: [false, false, true, false, false, false, false]
}, 
{
startTime: '11:00am',
endTime: '6:00pm',
open: [false, false, false, true, false, false, true]
}
];
// get open time of the store on d'th day. Returns null if the shop is closed on 
// that day.
function getOpenTime(d){
for(var i = 0; i < data.length; i++){
if(data[i].open[d]){
return {
startTime: data[i].startTime,
endTime: data[i].endTime
}
}
}
return null
}
// Helper function to format an `10:00am-5:00pm Monday-Wednesday` format, start and 
// end are index in the daysArr representing corresponding day of the week.
function dayFormatter(openTime, start, end) {
return openTime.startTime + '-' + openTime.endTime + ' ' + daysArr[start] + (start == end ? '' : '-' + daysArr[end])
}
function getOpenDays(){
result = []
for(var i = 0; i < 7; i++){
var iThDayOpenTime = getOpenTime(i);
if(iThDayOpenTime != null){
// If the shop is open on last day of the week, then no need to check any 
// further, and the open time should be pushed to the result.
if(i == 6){
result.push(dayFormatter(iThDayOpenTime, i, i));
}
for(var j = i + 1; j < 7; j++){
var jThDayOpenTime = getOpenTime(j);
// If the opentime of j'th day does not match with i'th day, that
// means all the days from i to j - 1, the shop is open on a same
// time, which is equal to open time of i'th day
if(jThDayOpenTime == null || iThDayOpenTime.startTime !== jThDayOpenTime.startTime || iThDayOpenTime.endTime !== jThDayOpenTime.endTime) {
result.push(dayFormatter(iThDayOpenTime, i, j - 1));
i = j - 1;
break;
}
// If the opentime of j'th day matches i'th day, there is only 
// one case when j == 6, that means we have no other days to 
// look at. So we know that all days from i to j, then shop is
// open on a same time.
else {
if(j == 6){
result.push(dayFormatter(iThDayOpenTime, i, j));
i = j;
}
}
}
}
}
return result
}
console.log(getOpenDays());