合并多个数组到另一个数组 [英] Combine multiple array to another array

问题描述

我想要像这样组合2个数组

i want combine 2 array like this

示例1：

Arr1 = ['A','B','C'],
Arr2 = ['D','E']

将成为

Arr3 = [
  ['A','D'],['A','E'],['B','D'],['B','E'],['',''] ....
]

示例2：

Arr1 = ['A','B','C'],
Arr2 = ['D','E']
Arr3 = ['G','H']

将成为

Arr4 = [
  ['A','D','G'],['A','E','G'],['','',''].... 
]

任何想法或建议我的算法都可以如此感谢

Any idea or suggest me algorithm can be like this thanks so much

推荐答案

您建议的问题与密码挑战。这个想法是使用回溯

The problem you suggest is very similar to this Leetcode Challenge. The idea is to use Backtracking

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 

所有的计算，可以像 CartesianProduct（0，要乘以Array的列表，）

假设您的 ArrayList = [['A']，['C'，'D']] 。而 CP 是 CartesianProduct

                 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（但在C ++中）。希望它能给您一些想法，用PHP编写

My solution to the Leetcode problem(but in C++). Hopefully it'll give you some idea to write in PHP

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;
    }
};

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