Question

假设我有像

这样的数字数组

00
05
10
15
upto
55

现在我想开发一个例程，我将传递数组和输入数字。假设我将32作为输入传递，因此将搜索32 数组和返回35因为35是最接近最高的32，所以只需指导我如何像那个例行程一样发展。

这里我开发了一个返回30最接近32但我想要35应该返回。我知道常规中存在一些缺陷。如果有人帮我构建正确的例行程序，那会有所帮助。

这是我的，没有返回预期的结果。

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript" >
$(document).ready(function(){
            var array = [00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55];
            var number = 32;
            alert (closest (number, array)); // display output here

        function closest (num, arr) {
                var curr = arr[0];
                var diff = Math.abs (num - curr);
                for (var val = 0; val < arr.length; val++) {
                    var newdiff = Math.abs (num - arr[val]);
                    if (newdiff < diff) {
                        diff = newdiff;
                        curr = arr[val];
                    }
                }
                curr=(curr>=10)?curr:'0'+curr;  //If result is 5, to display '05' instead of '5'
                return curr;
            }
});
</script>

感谢

我会接受这个

$(document).ready(function () {
        var array = [00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55];
        test(32)
        test(4)
        test(100)
        test(47)
        test(45)

        function test(num) {
            //log(num + ' --> ' + closest(num, array)); // display output here
            alert(closest(num, array));
        }

        function closest(num, arr) {
            var curr = Infinity;
            for (var i = 0; i < arr.length; i++) {
                if (arr[i] >= num && curr > arr[i]) {
                    curr = arr[i]
                }
            }
            curr = (curr >= 10) ? curr : '0' + curr; //If result is 5, to display '05' instead of '5'
            return curr;
        }
    });

    var log = (function () {
        var $log = $('#log');
        return function (msg) {
            $('<p/>', {
                text: msg
            }).prependTo($log)
        }
    })();

Answer 1

你可以做到

＆＃13;

$(document).ready(function() {
  var array = [00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55];
  test(32)
  test(4)
  test(100)
  test(47)
  test(45)

  function test(num) {
    log(num + ' --> ' + closest(num, array)); // display output here
  }

  function closest(num, arr) {
    var curr = Infinity;
    for (var i = 0; i < arr.length; i++) {
      if (arr[i] >= num && curr > arr[i]) {
        curr = arr[i]
      }
    }
    curr = (curr >= 10) ? curr : '0' + curr; //If result is 5, to display '05' instead of '5'
    return curr;
  }
});

var log = (function() {
  var $log = $('#log');
  return function(msg) {
    $('<p/>', {
      text: msg
    }).prependTo($log)
  }
})();

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

<div id="log"></div>

＆＃13;

Answer 2

快(O(n))并且足够简单。

此方法获得精确且最接近的值。低，精确和高。

var max = function (arr) { return  Math.max.apply(null, arr); };
var min = function (arr) { return  Math.min.apply(null, arr); };
    
var nearest  = function (arr, x) {
    var l = [], h = [];
    var exact = -1;
    
    arr.forEach(function (v) {           
       if (v == x) exact = v;
       ((v < x) && l.push(v)) || ((v > x) && h.push(v));
    });
       
    return { "low": arr.indexOf(max(l)), "exact": exact, "high": arr.indexOf(min(h)) };
};

var array = [00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55];
 
var testValue = 30;

var range = nearest(array, testValue);

if (range.exact != -1) {
     alert("Exact Value " + testValue) 
}
alert("Near Low:" + array[range.low] + " Near High:" + array[range.high]);

Answer 3

http://jsfiddle.net/59n6f5qv/1/

<强>更新

var array = [00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55];
var number = 32;
alert(closest(number, array)); // display output here


function closest(num, arr) {    
    var next = Math.max.apply(Math, arr);


    for (var val = 0; val < arr.length; val++) {
        if (arr[val] >= number && arr[val] < next) next = arr[val];        
    }    

    curr = next;

    curr = (curr >= 10) ? curr : '0' + curr; //If result is 5, to display '05' instead of '5'
    return curr;
}

Answer 4

function closest(num, arr) {
    if (arr.length == 0) { throw ("Array is empty"); }
    if (arr[0] > actualnum) { return arr[0]; }
    var diff = actualnum - arr[0];
    for (var i = 1; i < arr.length; i++) {
        if (actualnum <= arr[i]) {
            return (arr[i] - actualnum > diff) ? arr[i - 1] : arr[i];
        } else {
        diff = actualnum - arr[i];
        }
    }
    return arr[arr.length - 1];
}

如何计算范围内最接近的最近最近数

我会接受这个

4 个答案: