為什么通過 char 向量進行迭代比通過 C++ 中的字符串進行迭代更快

Question

我正在練習編碼挑戰，我必須reverse the vowels in a string 。

My first approach failed because of exeeding Time limit. 這是我使用字符串迭代來反轉字符串中的元音的第一種方法。

string reverseVowels(string s) {
    string str = "";
    //storing the vowels from the string into another string
    for (auto x : s)
        if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U')
            str = str + x;
    //swapping the vowels
    int count = 0;
    for (int i = s.size() - 1; i >= 0; i--)
    {
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I' || s[i] == 'O' || s[i] == 'U')
        {
            s[i] = str[count];
            count++;
        }
    }
        return s;
    }

My second approach using the char vector iteration had passed all the tests. 這是我的第二種方法

class Solution {
public:
    string reverseVowels(string s) {
  vector<char> v;
  //storing the vowels from the string into vector
    for (auto x : s)
        if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U')
            v.push_back(x);
    //swapping the vowels
    int count = 0;
    for (int i = s.size() - 1; i >= 0; i--)
    {
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I' || s[i] == 'O' || s[i] == 'U')
        {
            s[i] = v[count];
            count++;
        }
    }
        return s;
    }
};

你能解釋一下為什么我的第一種方法沒有通過測試，而第二種方法通過了測試

Answer 1

替換str = str + x; 與str.push_back(x); 或str += x; ，您可能會看到與vector相同的性能。

str = str + x; 制作str的副本，將字符附加到該副本，然后在分配回str時制作另一個副本。 結果，您的算法是二次的，沒有充分的理由。

Answer 2

這是因為您正在執行str = str + x ，這會創建str的不必要副本，但是std::vector::push_back或std::string::push_back將字符附加到向量或字符串，這比創建str的副本。

Answer 3

str = str + x這會在復制時創建一個額外的 str 副本。 std::vector::push_back這個直接附加到向量字符串