LeetCode-345. Reverse vowels in a string

description

Write a function that takes a string as input and reverses the vowels in the string.

 

Example 1:

Input: "hello"
Output: "holle"
Example 2:

Input: "leetcode"
Output: "leotcede"
 

prompt:

Vowels do not contain the letter "y".

Source: LeetCode
Link: https://leetcode-cn.com/problems/reverse-vowels-of-a-string/

Solve

    class Solution {
    private:
        std::unordered_set<char> vowelsRec{'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};

    public:
        // 方法一，申请额外的空间存储元音字母，时间复杂度O(N), 空间复杂度O(K),其中K < N
        string reverseVowels_1e(string s) {
            string strVowels;
            const int n = s.size();
            for (int i = 0; i < n; ++i) {
                char lc = tolower(s[i]);  //转换为小写字母确定是否是元音字母，减少条件判断
                if (lc == 'a' || lc == 'e' || lc == 'i' || lc == 'o' || lc == 'u') {
                    strVowels.push_back(s[i]);
                }
            }
            std::reverse(strVowels.begin(), strVowels.end());

            int k = 0;
            for (int i = 0; i < n; ++i) {
                char lc = tolower(s[i]);
                if (lc == 'a' || lc == 'e' || lc == 'i' || lc == 'o' || lc == 'u') {
                    s[i] = strVowels[k++];
                }
            }
            return s;
        }

        // 方法二，指针对撞，直接原地修改，时间复杂度O(N), 空间复杂度O(1)
        string reverseVowels_2e(string s) {
            const int n = s.size();
            int low = 0;
            int high = n - 1;
            while (low < high) {
                char llc = tolower(s[low]);
                while (low < n - 1 && (llc != 'a' && llc != 'e' && llc != 'i' && llc != 'o' && llc != 'u')) {
                    llc = tolower(s[++low]);
                }

                char lhc = tolower(s[high]);
                while (high > 0 && (lhc != 'a' && lhc != 'e' && lhc != 'i' && lhc != 'o' && lhc != 'u')) {
                    lhc = tolower(s[--high]);
                }

                if (low > high) {
                    return s;
                }

                std::swap(s[low], s[high]);
                ++low;
                --high;
            }
            return s;
        }

        // 方法二优化版，将元音字母的条件修改为查询，指针对撞，直接原地修改，时间复杂度O(N), 空间复杂度O(1)
        string reverseVowels(string s) {
            const int n = s.size();
            int low = 0;
            int high = n - 1;
            while (low < high) {
                while (low < n - 1 && (vowelsRec.count(s[low]) == 0)) {
                    ++low;
                }

                while (high > 0 && (vowelsRec.count(s[high]) == 0)) {
                    --high;
                }

                if (low > high) {
                    return s;
                }

                std::swap(s[low], s[high]);
                ++low;
                --high;
            }
            return s;
        }
    };