我有一组数字,例如[300, 500, 700, 1000, 2000, 3000]
,我想找到最接近的数字,而不是在给定的数字之下。
例如,搜索2200将返回3000(不是2000)。
但是,如果我搜索3200,因为数组中没有更高的值,它应该返回3000,因为没有其他选择。
我可以使用以下方法获得最接近该值的数字:
if (sizeToUse == null || Math.abs(this - monitorWidth) < Math.abs(sizeToUse - monitorWidth)) {
sizeToUse = this;
}
然而,我无法完成所有工作。我的完整代码是:
$(function() {
var monitorWidth = window.screen.availWidth,
sizeToUse = null,
upscaleImages = false;
$('.responsive-img').each(function(){
var sizeData = $(this).attr('data-available-sizes');
sizeData = sizeData.replace(' ', '');
var sizesAvailable = sizeData.split(',');
sizesAvailable.sort(function(a, b){return b-a});
$.each(sizesAvailable, function(){
if(upscaleImages){
if (sizeToUse == null || Math.abs(this - monitorWidth) < Math.abs(sizeToUse - monitorWidth)) {
sizeToUse = this;
}
}
else{
//We don't want to upscale images so we need to find the next highest image available
}
});
console.log('Size to use ' + sizeToUse + ' monitor width ' + monitorWidth);
});
});
答案 0 :(得分:6)
您可以使用此代码:
function closest(arr, closestTo){
var closest = Math.max.apply(null, arr); //Get the highest number in arr in case it match nothing.
for(var i = 0; i < arr.length; i++){ //Loop the array
if(arr[i] >= closestTo && arr[i] < closest) closest = arr[i]; //Check if it's higher than your number, but lower than your closest value
}
return closest; // return the value
}
var x = closest(yourArr, 2200);
答案 1 :(得分:1)
var list = [300, 500, 700, 1000, 2000, 3000];
function findBestMatch(toMatch) {
// Assumes the array is sorted.
var bestMatch = null;
var max = Number.MIN_VALUE;
var item;
for (var i = 0; i < list.length; i++) {
item = list[i];
if (item > toMatch) {
bestMatch = item;
break;
}
max = Math.max(max, item);
}
// Compare to null, just in case bestMatch is 0 itself.
if (bestMatch !== null) {
return bestMatch;
}
return max;
}
alert(findBestMatch(2200));
alert(findBestMatch(3200));
答案 2 :(得分:0)
sizesAvailable.sort(function(a, b){return a-b}); // DESCENDING sort
if(upscaleImages) // do th eif once, not every time through the loop
{
$.each(sizesAvailable, function()
{
if (this > monitorWidth)
sizeToUse = this;
}
if (sizeToUse == null) sizeToUse = sizesAvailable[0];
}
else
{
$.each(sizesAvailable, function()
{
//We don't want to upscale images so....
}
}
});
答案 3 :(得分:0)
另一种方法是找到第一个大于或等于你想要的候选,并取结果,如果没有匹配则返回最后一个元素:
function closestNumberOver(x, arr) {
return arr.find(d => d >= x) || arr[arr.length - 1]
}