通过置换四个给定数字找出最大可能时间HH:MM

Question

我最近在工作中进行了促销编码测试.这是我真正努力的任务之一,并且想知道最好的方法是什么.我使用if和if的负载,而不是最干净的解决方案,但完成了工作.

我被问到的问题是:

将4个数字格式化为24小时(00:00),找到可能的最长(最晚)时间,考虑到最大小时数为23,最大分钟数为59.如果不可能,则返回NOT POSSIBLE.

例如:

6,5,2,0将是20:56

3,9,5,0将是09:53

7,6,3,8是不可能的

必须返回时间或字符串的示例函数看起来像这样,A,B,C,D与上面以逗号分隔的列表不同的数字:

function generate(A, B, C, D) {
    // Your code here
} 

人们将如何解决这个问题？

Answer 1

这是我提出的非暴力解决方案.查看代码中的注释,了解它是如何工作的.如果其中任何一个不清楚我可以帮助澄清.

function generate(A, B, C, D) {
    vals = [A, B, C, D];
    counts = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
    for (i = 0; i < vals.length; i++) {
        for (j = vals[i]; j < counts.length; j++) counts[j]++;
    }
    // counts is now populated with the number of values less than or equal to the index it belongs to
    // so counts[2] is the total number of 0's, 1's and 2's
    if (counts[2] === 0) return 'NOT POSSIBLE';
    // if there are no 0's and 1's, then it must start with 2
    mustStartWith2 = counts[1] === 0;
    if (mustStartWith2 && counts[3] === 1) return 'NOT POSSIBLE';
    // We want a count of the number of free digits that are 5 or less (for the minute digit)
    numbersAvailableForMinute = counts[5] - (mustStartWith2 ? 2 : 1); 
    if (numbersAvailableForMinute === 0) return 'NOT POSSIBLE';
    // we now know that it is a valid time
    time = [0, 0, 0, 0];
    // we also know if it starts with 2
    startsWith2 = mustStartWith2 || (numbersAvailableForMinute >= 2 && counts[2] > counts[1]);
    // knowing the starting digit, we know the maximum value for each digit
    maxs = startsWith2 ? [2, 3, 5, 9] : [1, 9, 5, 9];
    for (i = 0; i < maxs.length; i++) {
        // find the first occurrence in counts that has the same count as the maximum
        time[i] = counts.indexOf(counts[maxs[i]]);
        // update counts after the value was removed
        for (j = time[i]; j < counts.length; j++) counts[j]--;
    }
    // create the time
    return time[0]+""+time[1]+":"+time[2]+""+time[3];
}

Answer 2

一旦我了解到您可以将问题视为“生成一个小于 24 的数字和一个小于 60 的数字”而不是尝试处理单个数字，这一事实的推理就变得容易多了。

这将遍历集合中的数字对，找到可以从该数字对中得出的最大有效小时，然后找到可以从剩余数字中得出的最大有效分钟。

var generate = function(a, b, c, d) {
  var biggest = function(a, b, max) {
    // returns largest of 'ab' or 'ba' which is below max, or false.
    // I'm sure there's a more concise way to do this, but:
    var x = '' + a + b;
    var y = '' + b + a;
    if (max > x && max > y) {
      var tmp = Math.max(x,y);
      return (tmp < 10) ? "0"+tmp : tmp;
    }
    if (max > x) return x;
    if (max > y) return y;
    return false;
  }

  var output = false;

  var input = [].slice.call(arguments);
  for (var i = 0; i < arguments.length; i++) {
    for (var j = i + 1; j < arguments.length; j++) {
      // for every pair of numbers in the input:
      var hour = biggest(input[i], input[j], 24); // What's the biggest valid hour we can make of that pair?
      if (hour) {
        // do the leftovers make a valid minute?
        var tmp = input.slice(); // copy the input
        tmp.splice(j, 1);
        tmp.splice(i, 1);
        var minute = biggest(tmp[0], tmp[1], 60);
        if (hour && minute) {
          // keep this one if it's bigger than what we had before:
          var nval = hour + ':' + minute;
          if (!output || nval > output) output = nval;
        }
      }
    }
  }
  return output || 'NOT POSSIBLE';
}

/* --------------- Start correctness test --------------------- */
  var tests = ['0000', '1212', '1234', '2359', '2360','2362','2366', '1415', '1112', '1277', '9999', '0101'];
console.log('---');
for (var i = 0; i < tests.length; i++) {
  console.log(
    tests[i],
    generate.apply(this, tests[i].split(''))
  )
}



/* --------------- Start Speed Test --------------------- */

let startTime = Math.floor(Date.now());
let times = 10000; //how many generate call you want?
let timesHolder = times;

while (times--) {
  let A = randNum();
  let B = randNum();
  let C = randNum();
  let D = randNum();
  generate(A, B, C, D);
  if (times == 0) {
    let totalTime = Math.floor(Date.now()) - startTime;
    let msg = timesHolder + ' Call Finished Within -> ' + totalTime + ' ms <-';
    console.log(msg);
    // alert(msg);
  }
}

function randNum() {
  return Math.floor(Math.random() * (9 - 0 + 1)) + 0;
}

/* --------------- END Speed Test --------------------- */