相等地填充数组

Filling arrays equally

本文关键字：数组 填充      更新时间：2023-09-26

给定两个数组(最大长度为15):

var a = [1,2,3];
var b = [5,6,7,8,9,10,11,12];
var data = [13,14,15,16,17,18,19,20,21,22,23,24,25,26];

用data数组的数据"填充"数组a和b的最佳方法是什么，因此它们都是相同的长度(或接近它)，但不超过最大长度?

要求的结果可以看起来像这样:

var a = [1,2,3,13,14,15,16,17,18,19,20,21];
var b = [5,6,7,8,9,10,11,12,22,23,24,25,26];

我的情况更复杂，我正在寻找一种有效的方法:我在屏幕上有2面，每面可能有来自FB的ppl的个人资料照片。如果没有足够的照片在一边，我需要添加"假"照片来填充它，所以两边看起来或多或少会均匀地填充照片

var a    = [1,2,3];
var b    = [5,6,7,8,9,10,11,12];
var data = [13,14,15,16,17,18,19,20,21,22,23,24,25,26];
for( var i = data.length; i--; ){
  if (a.length > b.length)
      b.push(data[i])
  else
      a.push(data[i])
}
console.log(a.length, b.length);

这里有另一种不需要遍历元素的方法:

var a = [1,2,3],
    b = [5,6,7,8,9,10,11,12],
    data = [13,14,15,16,17,18,19,20,21,22,23,24,25,26];
var targetLength = (a.length + b.length + data.length)/2,
    aCnt = Math.ceil(targetLength) - a.length,
    bCnt = Math.floor(targetLength) - b.length;
a = a.concat(data.slice(0, aCnt));
b = b.concat(data.slice(aCnt, data.length));
// a is [1, 2, 3, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22]
// b is [5, 6, 7, 8, 9, 10, 11, 12, 23, 24, 25, 26]

从Brad的评论来看，我认为你可以这样做:

ab=[];
ab=ab.concat(a,b,data);
del = (a.length + b.length + data.length) / 2;
a = [];
b = [];
for (var i = 0; i < Math.ceil(del); i++) {
    a.push(ab[i]);
    b.push(ab[Math.ceil(del) + i]);
}

算法如下:

1)识别a中元素的个数

2)确定b中元素的个数

3)确定data中元素的个数

4)如果a中的元素小于b，则从data填充a，直到它等于b，反之亦然

5)如果a和b中的元素相等，并且data中仍然有元素，则依次填充数组a和b。

你想让a和b的最后一个索引相差12，最后一个数组的len是12!!我认为你可以使用这个算法!: 

for j in data:
  while ( j-max(a)) == 12 ):
   x++
for i in range(x):
 data[i]=a[i]

，然后删除索引号大于12的索引!

我的解决方案是:

var a    = [1,2,3];
var b    = [5,6,7,8,9,10,11,12];
var data = [13,14,15,16,17,18,19,20,21,22,23,24,25,26];
var howManyToFill = (data.length + (b.length + a.length)/2)/2;
// take half of the "data" and put it in "a"
a = a.concat(data.slice(0,howManyToFill));
// what's left goes to "b"
b = b.concat(data.slice(howManyToFill));
console.log(a);
console.log(b);

var b = [1,2,3];
var a = [5,6,7,8,9,10,11,12];
var data = [13,14,15,16,17,18,19,20,21,22,23,24,25,26];
// find how many more elements need to be added to the smaller array 
var diff = Math.abs(b.length - a.length);
// get the elements to add to the smallest array
var data1 = data.splice(0, diff + ((data.length - diff) / 2));
// data now only contains the elements that will not be added to the smallest array
if (a.length < b.length) {
    // a is the smallest array
    a.push.apply(a,  data1);
    b.push.apply(b, data);
} else {
    // b is the smallest array
    b.push.apply(b,  data1);
    a.push.apply(a, data);
}
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