如何根据数组值计算空闲可用时间

Question

假设我有一个如下所示的数组：

{
    meeting: 'one',
    start_time: "07:15:00",
    end_time: "08:15:00"
},
{
    meeting: 'two',
    start_time: "15:45:00",
    end_time: "18:15:00"
}

现在我想计算我从 00:00:00 到 24:00:00 的所有空闲时间。

例如，新数组将如下所示：

{
    meeting: 'free time',
    start_time: "00:00:00",
    end_time: "07:14:00"
},
{
    meeting: 'one',
    start_time: "07:15:00",
    end_time: "08:15:00"
},
{
    meeting: 'free time',
    start_time: "08:16:00",
    end_time: "15:44:00"
},
{
    meeting: 'two',
    start_time: "15:45:00",
    end_time: "18:15:00"
}

关于如何在没有插件的情况下完成此操作的任何建议？

Answer 1

This is what I'll do:

function meeting(meeting, time1, time2) {
    this.meeting = meeting;
    this.start_time = time1;
    this.end_time = time2;
}


meeting.prototype.convert = function (a) {
    var b = this[a + '_time'];
    var start = b.slice(0, 3);
    var min = parseInt(b.slice(3, 5));
    min = a == "start" ? min - 1 : min + 1;
    min = min < 10 ? "0" + min : min;
    var ending = b.slice(5, 8);

    return start + min + ending

}

Array.prototype.getFreeTime = function () {
    var l = this.length,
        withFreeTime = [],
        i = 0;
    while (i < l) {

        var s = this[i - 1] ? this[i - 1].convert('end') : "00:00:00";
        withFreeTime.push(new meeting('freetime', s, this[i].convert('start')));
        withFreeTime.push(this[i]);
        i++;
    }

    //to add the freetime from "18:16:00" to  "23:59:00":
    withFreeTime.push(new meeting('freetime', this[i-1].convert('end'),'23:59:00'));

    return withFreeTime;
}

var meetings = [];
meetings[0] = new meeting('meeting1', "07:15:00", "08:15:00");
meetings[1] = new meeting('meeting2', "15:45:00", "18:15:00");
meetings = meetings.getFreeTime();
console.log(JSON.stringify(meetings));

consloe.log：

[{"meeting":"freetime","start_time":"00:00:00","end_time":"07:14:00"},
{"meeting":"meeting1","start_time":"07:15:00","end_time":"08:15:00"},
{"meeting":"freetime","start_time":"08:16:00","end_time":"15:44:00"},
{"meeting":"meeting2","start_time":"15:45:00","end_time":"18:15:00"},
{"meeting":"freetime","start_time":"18:16:00","end_time":"23:59:00"}]

Answer 2

您需要在跟踪时间的同时遍历数组。 jsFiddle

function subtractMinute(time){
  var h = +time.substr(0, 2);
  var m = +time.substr(3, 2);

  if(m > 0){
    m -= 1;
  }else{
    if(h > 0){
      h -= 1;
    }else{
      return false;
    }
    m = 59;
  }

  if(h < 10)
    h = '0'+h;

  if(m < 10)
    m = '0'+m;

  return h+':'+m+':00';
}

function addMinute(time){
  var h = +time.substr(0, 2);
  var m = +time.substr(3, 2);

  if(m < 59){
    m += 1;
  }else{
    if(h < 22){
      h += 1;
    }else{
      return false;
    }
    m = 0;
  }

  if(h < 10)
    h = '0'+h;

  if(m < 10)
    m = '0'+m;

  return h+':'+m+':00';
}

//If meetings is already sorted by time
//you can skip this next bit of code
meetings.sort(function(a, b){
  return a.start_time > b.start_time? 1: -1;
});

var schedule = [];
var start_time = '00:00:00';
var end_time = '23:59:00';
for(var i=0, l=meetings.length; i<l; i++){
  end_time = subtractMinute(meetings[i].start_time);

  if(i)
    start_time = addMinute(meetings[i-1].end_time);

  if((end_time && !i) || (end_time && i && meetings[i-1].end_time < meetings[i].start_time))
    schedule.push({meeting: 'free time', start_time: start_time, end_time: end_time});

  schedule.push(meetings[i]);

  if(i+1 === l){
    start_time = addMinute(meetings[i].end_time);

    if(start_time)
        schedule.push({meeting: 'free time', start_time: start_time, end_time: '23:59:00'});
  }
}

console.log(schedule);

Answer 3

只是一个简单的例子

var schedule = [{
    meeting: 'free time',
    start_time: "00:00:00",
    end_time: "07:14:00"
},
{
    meeting: 'one',
    start_time: "07:15:00",
    end_time: "08:15:00"
},
{
    meeting: 'free time',
    start_time: "08:16:00",
    end_time: "15:44:00"
},
{
    meeting: 'two',
    start_time: "15:45:00",
    end_time: "18:15:00"
}]

var st,
    se,
    fh, fm, fs,
    ft = 0;

for (var i = 0; i < schedule.length; i++) {

    var s = schedule[i];

    if (s['meeting'] === 'free time') {

        st = s['start_time'].replace(/(\d+):(\d+):(\d+)/, function(t,h,m,s) {
          return (h * 3600) + (m * 60) + (+s);  
          // s is a string, +s converts it to an integer

        })            
        et = s['end_time'].replace(/(\d+):(\d+):(\d+)/, function(t,h,m,s) {
          return (h * 3600) + (m * 60) + (+s);         
        })         

        /* your free time in seconds for this slot */
        ft += (et-st);
    }
}

fh = ~~(ft / 3600);
fm = ~~((ft - (fh * 3600)) / 60);
fs = ft - (fh * 3600) - (fm * 60)

console.log("you have %s:%s:%s free time", fh, fm, fs);
// you have 14:42:0 free time

Answer 4

这是一种可能的解决方案 (JSFiddle here)：

var time = [{
    meeting: 'one',
    start_time: "07:15:00",
    end_time: "08:15:00"
}, {
    meeting: 'two',
    start_time: "15:45:00",
    end_time: "18:15:00"
}];

// out contains the result
// timeline is a list of minutes containing some event
var out = [], timeline = [];

// Add free time after the last meeting:
// timeline[23*60] = undefined;

// fill up the timeline
for (var m = 0, current = 0; m < 24 * 60 && current < time.length; m += 1) {
    var start = str_as_min(time[current].start_time);
    var end = str_as_min(time[current].end_time);

    if (m >= start) {
        if (m <= end) {
            timeline[m] = time[current].meeting;
        } else {
            current += 1;
        }
    }
}

// fill up the out
for (var m = 0, last = null; m < timeline.length; m += 1) {
    var current = timeline[m];

    if (current !== last) {
        if (out.length > 0) {
            out[out.length-1].end_time = min_as_str(m-1);
        }

        out.push({
            start_time: min_as_str(m),
            meeting: timeline[m] === undefined ? 'free_time' : timeline[m]
        });
    }

    if (m === timeline.length - 1) {
        out[out.length-1].end_time = min_as_str(m);
    }

    last = current;
}

console.log(JSON.stringify(out));

// helper functions
function min_as_str (min) {
    return Math.floor(min / 60) + ':' + (min % 60);
}

function str_as_min (str) {
    var m = str.match(/^(\d\d):(\d\d)/);
    return 60 * parseInt(m[1], 10) + parseInt(m[2], 10);
}

与您建议的示例输出数组的唯一区别是时间格式（即7:15 而不是07:15 - 您可以轻松解决此问题）。