Php 将多个阵列合并到另一个阵列
我想要这样的组合2数组 例1:Php 将多个阵列合并到另一个阵列,php,algorithm,Php,Algorithm,我想要这样的组合2数组 例1: Arr1 = ['A','B','C'], Arr2 = ['D','E'] 将成为 Arr3 = [ ['A','D'],['A','E'],['B','D'],['B','E'],['',''] .... ] Arr4 = [ ['A','D','G'],['A','E','G'],['','',''].... ] 例2: Arr1 = ['A','B','C'], Arr2 = ['D','E'] Arr3 = ['G','H'] 将成为
Arr1 = ['A','B','C'],
Arr2 = ['D','E']
将成为
Arr3 = [
['A','D'],['A','E'],['B','D'],['B','E'],['',''] ....
]
Arr4 = [
['A','D','G'],['A','E','G'],['','','']....
]
例2:
Arr1 = ['A','B','C'],
Arr2 = ['D','E']
Arr3 = ['G','H']
将成为
Arr3 = [
['A','D'],['A','E'],['B','D'],['B','E'],['',''] ....
]
Arr4 = [
['A','D','G'],['A','E','G'],['','','']....
]
任何想法或建议我的算法都可以像这样,非常感谢简化版;按索引合并数组
$arr1 = range('a', 'b');
$arr2 = range('c', 'f');
$arr3 = range('g', 'k');
$arr4 = range('x', 'z');
$res = array();
// $counter = 1;
// while ($counter <= 4) {
// $array = "arr{$counter}";
// funcIndexMerge($res, $$array);
// $counter++;
// }
funcIndexMerge($res, $arr1);
funcIndexMerge($res, $arr2);
funcIndexMerge($res, $arr3);
funcIndexMerge($res, $arr4);
var_export($res);
function funcIndexMerge(&$res, $array) {
foreach ($array as $ari => $val) {
if (!isset($res[$ari])) {
$res[$ari] = array();
}
$res[$ari] = array_merge($res[$ari], array($val));
}
}
$arr1=范围('a','b');
$arr2=范围('c','f');
$arr3=范围('g','k');
$arr4=范围('x','z');
$res=array();
//$counter=1;
//while($counter$val){
如果(!isset($res[$ari])){
$res[$ari]=array();
}
$res[$ari]=数组合并($res[$ari],数组($val));
}
}
简化版;按索引合并数组
$arr1 = range('a', 'b');
$arr2 = range('c', 'f');
$arr3 = range('g', 'k');
$arr4 = range('x', 'z');
$res = array();
// $counter = 1;
// while ($counter <= 4) {
// $array = "arr{$counter}";
// funcIndexMerge($res, $$array);
// $counter++;
// }
funcIndexMerge($res, $arr1);
funcIndexMerge($res, $arr2);
funcIndexMerge($res, $arr3);
funcIndexMerge($res, $arr4);
var_export($res);
function funcIndexMerge(&$res, $array) {
foreach ($array as $ari => $val) {
if (!isset($res[$ari])) {
$res[$ari] = array();
}
$res[$ari] = array_merge($res[$ari], array($val));
}
}
$arr1=范围('a','b');
$arr2=范围('c','f');
$arr3=范围('g','k');
$arr4=范围('x','z');
$res=array();
//$counter=1;
//while($counter$val){
如果(!isset($res[$ari])){
$res[$ari]=array();
}
$res[$ari]=数组合并($res[$ari],数组($val));
}
}
您建议的问题与此非常相似。想法是使用
Psuedo PsedoCode:
result := string Array
CartesianProduct(rowIndex, ArrayList, stringSoFar):
if rowIndex equal ArrayList.size:
Add stringSoFar to result
return
for(eachItem in ArrayList[rowIndex])
CartesianProduct(rowIndex +1 , ArrayList, stringSoFar + eachItem)
return
这是一个执行所有计算的工作马,可以像CartesianProduct(0,要相乘的数组列表“”)一样调用它
假设您的ArrayList=[['A'],['C',D']]
。和CP
beCartesianProduct
CP(0, AL, "")
(Take 'A')/
/
CP(1, AL, "A") (stringSoFar becomes 'A')
(Take C) / \(Take 'D'.Second Iteration with second array['C', 'D'])
/ \
CP(2, AL, "AC") CP(2, AL, "AD")
/ \
rowIndex equal size. rowIndex equals listSize i.e No more list to look
Add "AC" and return Add stringsoFar ("AD") to result and rerturn
我对Leetcode问题的解决方案(但使用C++)。希望它能给你一些用PHP编写的想法
class Solution {
public:
map<char, vector<string>> values {
{'2', vector<string>{"a", "b", "c"}},
{'3', vector<string>{"d", "e", "f"}},
{'4', vector<string>{"g", "h", "i"}},
{'5', vector<string>{"j", "k", "l"}},
{'6', vector<string>{"m", "n", "o"}},
{'7', vector<string>{"p", "q", "r", "s"}},
{'8', vector<string>{"t", "u", "v"}},
{'9', vector<string>{"w", "x", "y", "z"}}
};
vector<string> answer;
void doComb(int index, string digits, string sofar)
{
if(index == digits.size())
{
if(sofar != "")
{
answer.push_back(sofar);
}
return;
}
for(auto lett : values[digits[index]])
{
doComb(index + 1, digits, sofar + lett);
}
return;
}
vector<string> letterCombinations(string digits) {
doComb(0, digits, "");
return answer;
}
};
类解决方案{
公众:
映射值{
{'2',向量{“a”,“b”,“c”},
{'3',向量{“d”,“e”,“f”},
{'4',向量{“g”,“h”,“i”},
{'5',向量{“j”,“k”,“l”},
{'6',向量{“m”,“n”,“o”},
{'7',向量{“p”,“q”,“r”,“s”},
{'8',向量{“t”,“u”,“v”},
{'9',向量{“w”,“x”,“y”,“z”}
};
向量答案;
void doComb(整数索引、字符串数字、字符串sofar)
{
如果(索引==位数.size())
{
如果(sofar!=“”)
{
回答:向后推(索法尔);
}
返回;
}
用于(自动:值[位数[索引])
{
文档(索引+1,数字,sofar+lett);
}
返回;
}
矢量字母组合(字符串数字){
doComb(0,数字,“”);
返回答案;
}
};
您建议的问题与此非常相似。想法是使用
Psuedo PsedoCode:
result := string Array
CartesianProduct(rowIndex, ArrayList, stringSoFar):
if rowIndex equal ArrayList.size:
Add stringSoFar to result
return
for(eachItem in ArrayList[rowIndex])
CartesianProduct(rowIndex +1 , ArrayList, stringSoFar + eachItem)
return
这是一个执行所有计算的工作马,可以像CartesianProduct(0,要相乘的数组列表“”)一样调用它
假设您的ArrayList=[['A'],['C',D']]
。和CP
beCartesianProduct
CP(0, AL, "")
(Take 'A')/
/
CP(1, AL, "A") (stringSoFar becomes 'A')
(Take C) / \(Take 'D'.Second Iteration with second array['C', 'D'])
/ \
CP(2, AL, "AC") CP(2, AL, "AD")
/ \
rowIndex equal size. rowIndex equals listSize i.e No more list to look
Add "AC" and return Add stringsoFar ("AD") to result and rerturn
我对Leetcode问题的解决方案(但使用C++)。希望它能给你一些用PHP编写的想法
class Solution {
public:
map<char, vector<string>> values {
{'2', vector<string>{"a", "b", "c"}},
{'3', vector<string>{"d", "e", "f"}},
{'4', vector<string>{"g", "h", "i"}},
{'5', vector<string>{"j", "k", "l"}},
{'6', vector<string>{"m", "n", "o"}},
{'7', vector<string>{"p", "q", "r", "s"}},
{'8', vector<string>{"t", "u", "v"}},
{'9', vector<string>{"w", "x", "y", "z"}}
};
vector<string> answer;
void doComb(int index, string digits, string sofar)
{
if(index == digits.size())
{
if(sofar != "")
{
answer.push_back(sofar);
}
return;
}
for(auto lett : values[digits[index]])
{
doComb(index + 1, digits, sofar + lett);
}
return;
}
vector<string> letterCombinations(string digits) {
doComb(0, digits, "");
return answer;
}
};
类解决方案{
公众:
映射值{
{'2',向量{“a”,“b”,“c”},
{'3',向量{“d”,“e”,“f”},
{'4',向量{“g”,“h”,“i”},
{'5',向量{“j”,“k”,“l”},
{'6',向量{“m”,“n”,“o”},
{'7',向量{“p”,“q”,“r”,“s”},
{'8',向量{“t”,“u”,“v”},
{'9',向量{“w”,“x”,“y”,“z”}
};
向量答案;
void doComb(整数索引、字符串数字、字符串sofar)
{
如果(索引==位数.size())
{
如果(sofar!=“”)
{
回答:向后推(索法尔);
}
返回;
}
用于(自动:值[位数[索引])
{
文档(索引+1,数字,sofar+lett);
}
返回;
}
矢量字母组合(字符串数字){
doComb(0,数字,“”);
返回答案;
}
};
两个foreach循环。您的第二个示例(也称为Ex1)不够详细,我们无法知道发生了什么。听起来您在寻找两个数组的笛卡尔乘积。也许能帮上忙?这是我的答案,非常感谢我的朋友@tugayacNo的担忧,很高兴它很有用:)为什么是php标签?两个foreach循环。您的第二个示例(也称为Ex1)不够详细,我们无法知道发生了什么。听起来您在寻找两个数组的笛卡尔乘积。也许能帮上忙?这是我的答案,非常感谢我的朋友@tugayacNo的担忧,很高兴它很有用:)为什么使用php标签?