我有两个数组，数组1是一个施主数组，它有一系列值（不一定是我的例子中的相等值）。数组2是期望的结果，并且将存储一系列具有来自数组1的值的子数组，其中每个子数组的总和不会超过25.如果确实如此，超出部分将被推送到数组2中的下一个索引，其中规则将是也适用。
捐赠者阵列（阵列1）：
$a1=array(10,10,10,10,10,10,10,10,10,10);
所需的输出（数组2）：
Array
(
[0] => 10,10,5
[1] => 5,10,10
[2] => 10,10,5
[3] => 5,10,10
)
这里是我尝试过的代码，但是出现错误：
注意：未定义的偏移量： 10 ...等。
$a1=array(10,10,10,10,10,10,10,10,10,10);
$a2=array();
$count=count($a1);
for($i=0;$i<$count;$i++){
$a2count=array_sum($a2);
if($a2count>25){
$i=$i+1;
$a2[$i]=$a1[$i];
}
else{
$a2[$i]=$a1[$i];
}
}
print_r($a2);
我不知道要实现什么逻辑并获得我正在寻找的结果。
答案 0 :(得分：2)
也许这样的事情对你有用。我会注意到它不仅仅是一个复制粘贴的答案。也许有人会对它有所了解以便在将来改进它：
function slitArray($a1,$num = 25)
{
# Used to store the difference when the value exceeds the max
$store = 0;
# Storage container that will be built using sums/diffs
$new = array();
# Loop through the main array
foreach($a1 as $value) {
# If the last key/value pair in our return array is an array
if(is_array(end($new)))
# Add up the values in that array
$sum = array_sum(current($new));
else
# If not array, no values have been stored yet
$sum = 0;
# This just gets the last key
$count = (count($new)-1);
# Assign last key
$i = ($count <= 0)? 0 : $count;
# If the sum of the current storage array plus the value
# of the current array is greater than our max value
if(($sum + $value) > $num) {
# Take max and remove the current total of storage array
$use = ($num-$sum);
# Take what's left and remove it from the current value
$store = ($value-$use);
# If the current stored value (the value we want to push to
# the next storage k/v pair) is more than the max allowed
if($store > $num) {
# Takes a value, checks if it's greater than max,
# and if it is, breaks the value up by max as a divider
$divide = function($store,$num)
{
if($store > $num) {
$count = ceil($store/$num);
for($i=0; $i<$count; $i++) {
$new[] = ($store > $num)? $num : $store;
$store -= $num;
}
return $new;
}
else
return array($store);
};
# This should either be an array with 1 or more values
$forward = $divide($store,$num);
# Do a look forward and add this excess array into our
# current storage array
$a = $i;
foreach($forward as $aVal) {
$new[$a+=1][] = $aVal;
}
}
# If the store value is less than our max value, just add
# it to the next key in this storage array
else {
$new[$i+1][] = $store;
# Reset the storage back to 0, just incase
$store = 0;
}
}
# Set the current "use" value as the current value in our
# from-array. Since it doesn't exceed the max, it just gets
# added to the storage array
else
$use = $value;
# Sometimes the math makes $use 0, keep that out of the
# storage array. The $use value is the current value to add at
# the time of iteration. Previous storage values are added as
# future-keys
if($use > 0)
$new[$i][] = $use;
}
# Return the final assembled array
return $new;
}
# To use, add array into function
$a1 = array(10,10,10,10,10,10,10,10,10,10);
# to split using different max value, just add it to second arg
# example: slitArray($a1,20);
print_r(slitArray($a1));
给你：
Array
(
[0] => Array
(
[0] => 10
[1] => 10
[2] => 5
)
[1] => Array
(
[0] => 5
[1] => 10
[2] => 10
)
[2] => Array
(
[0] => 10
[1] => 10
[2] => 5
)
[3] => Array
(
[0] => 5
[1] => 10
[2] => 10
)
)
数组输入：
$a1 = array(23,2,71,23,50,2,3,4,1,2,50,75);
给你：
Array
(
[0] => Array
(
[0] => 23
[1] => 2
)
[1] => Array
(
[0] => 25
)
[2] => Array
(
[0] => 25
)
[3] => Array
(
[0] => 21
[1] => 4
)
[4] => Array
(
[0] => 19
[1] => 6
)
[5] => Array
(
[0] => 25
)
[6] => Array
(
[0] => 19
[1] => 2
[2] => 3
[3] => 1
)
[7] => Array
(
[0] => 3
[1] => 1
[2] => 2
[3] => 19
)
[8] => Array
(
[0] => 25
)
[9] => Array
(
[0] => 6
[1] => 19
)
[10] => Array
(
[0] => 25
)
[11] => Array
(
[0] => 25
)
[12] => Array
(
[0] => 6
)
)
答案 1 :(得分：1)
你走了：逻辑并不那么难。希望它有所帮助。
<?php
$a1=array(10,10,10,10,10,10,10,10,10,10);
$a2 = [];
$a3 = [];
$m = 0;
for($i = 0; $i < count($a1); ++$i){
$m += $a1[$i];
if($m > 25){
$n = $m % 25;
if(array_sum($a2) != 25){
$a2[] = $n;
}
$a3[] = implode(',', $a2);
$a2 = [];
$m = $n;
$a2[] = $n;
} else{
$a2[] = $a1[$i];
}
}
$a3[] = implode(',', $a2);
print_r($a3);
?>
答案 2 :(得分：0)
让我帮你使用Pseudocode：
ar1 = {10,10,10,20,40,[0]=>1,[0]=>3,[0]=>4};
ar2 = new array (ar.length) \\ worst case
int c = 0; \\current
foreach (ar1 as $value){
ar2 [c]+=ar1[i];
if (ar2 [c]>25){ c++;}
}
代码背后的逻辑：
Add ar1[i] to当前ar2值的值，直到它通过your limit（在这种情况下为25）。 If它超出了你的边界，than移动到目标数组中的next值。 worst case将是每个值超过25，因此它将是原始数组的exact copy。
这里是php代码：
$ar1=array(10,10,10,10,10,10,10,10,10,10);
$ar2 = array(0,0,0,0,0,0,0,0,0,0);
$c = 0;
foreach( $ar1 as $key => $value ){
$ar2[$c]=$value+$ar2[$c];
if ($ar2[$c]>25){$c++;}
}
答案 3 :(得分：0)
此问题的最终代码
<?php
function slitArray($a1,$num = 25)
{
$store = 0;
$new = array();
foreach($a1 as $value) {
if(is_array(end($new)))
$sum = array_sum(current($new));
else
$sum = 0;
$count = (count($new)-1);
$i = ($count <= 0)? 0 : $count;
if(($sum + $value) > $num) {
$use = ($num-$sum);
$store = ($value-$use);
if($store > $num) {
$divide = function($store,$num)
{
if($store > $num) {
$count = ceil($store/$num);
for($i=0; $i<$count; $i++) {
$new[] = ($store > $num)? $num : $store;
$store -= $num;
}
return $new;
}
else
return array($store);
};
$forward = $divide($store,$num);
$a = $i;
foreach($forward as $aVal) {
$new[$a+=1][] = $aVal;
}
}
else {
$new[$i+1][] = $store;
$store = 0;
}
}
else
$use = $value;
if($use > 0)
$new[$i][] = $use;
}
return $new;
}
$a1 = array(10,20,30,40,50,60);
$arr=slitArray($a1);
print_r($arr);
?>