如何检查JavaScript数组中是否存在某个值，继续添加0.15直到它不存在，然后将该值添加到数组的末尾？

Question

I have a number of text input elements that will contain the start time for a persons shift (Eg 9:00, 9:30, 10:00, etc).

My code will go through these elements one by one and push them to an array. If the value already exists in the array, I need to add 15 minutes (0.15) to the time and then check if it exists again. This will continue until the value does not exist and then push it to the array.

For example: John - 9:00 Bill - 9:00 Julie - 9:30 Sam - 9:30 Tony - 9:30

times = []

The first time is 9:00 which is not in times, so it will be added. Times is now equal to [9]. Next one is 9:00 which does exist, so I add 0.15, giving me 9.15 which does not exist, so now times = [9, 9.15]. This continues for all of the elements. If the decimal points equal .60 then I need to add 1 to the hour, so for Tony, the time added would be 10.

I know that I can just use an if statement to check this once, but how do I continue to add 15 and continue to check 2, 3, 4+ times?

if( times.contains(value)) {
      times.push(value + 0.15)
 }

To do it this way I would need heaps of if statements. How can I repeatedly check and then end the loop as soon as it's added?

Answer 1

You could take a Set and increment the value until it is not in the set, then map this value.

This approach uses a decimal value of 15 minutes, which is a quarter 0.25 hour.

 var times = [9, 9, 9.5, 9.5, 9.5], adjusted = times.map((s => t => { while (s.has(t)) { t += 0.25; } s.add(t); return t; })(new Set)); console.log(adjusted); 

Answer 2

One solution I can think of is

1. Store all the data in times regardless of their repeated value. eg [9:00,9:00,9:30,9:30,9:30];
2. Iterate data and if i'th data === i-1'th data increment by 15 min.

times = Arrays.sort(times);
for(let i=1;i<times.length;i++){
  times[i] = times[i]===times[i-1] ? add15min(times[i]) : times[i];
}

Iterations -:

1. i=1 (times[i] === times[i-1] so increment) -- [9:00,9:15,9:30,9:30,9:30];
2. i=2 (times[i] !== times[i-1] so no operation) -- [9:00,9:15,9:30,9:30,9:30];
3. i=3 (times[i] === times[i-1] so increment) -- [9:00,9:15,9:30,9:45,9:30];

4. i=4 (times[i] === times[i-1] so increment) -- [9:00,9:15,9:30,9:45,10:00];

   Final data set [9:00,9:15,9:30,9:45,10:00];